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Can this function be rewritten with a regex?
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I want to reformat and validate if a user has provided a valid Belgian enterprise number. Because the input can be all of the following examples:
- BE 0123.321.123
- BE0123.321.123
- BE0123 321 123
- 0123.321.123
- 123.321.123
- 123321123
I've written a function that validates and reformat the input to a 'display' version (BE 0123.123.123) and a 'code' version (123123123). This function looks like this.
formatAndValidateEnterpriseNumber = enterpriseNumber => {
if(enterpriseNumber === undefined || !enterpriseNumber || (enterpriseNumber || '').length < 3) return { isValid: false, error: 'Please fill in your enterprise number' };
//Remove space, dots, ...
enterpriseNumber = enterpriseNumber.toUpperCase();
enterpriseNumber = enterpriseNumber.replace(/[. ,:-]+/g, '');
//Check for double country code
const reDouble = /^[a-zA-Z]{4}/;
if (reDouble.test(enterpriseNumber)) enterpriseNumber = enterpriseNumber.substring(2);
if (enterpriseNumber.length < 9 || enterpriseNumber.length > 12) return { isValid: false, error: 'The length of the provided number is incorrect' };
//Check country code
const reBE = /^[a-zA-Z]{2}/;
if (reBE.test(enterpriseNumber)) {
//Check if country code = BE
if (enterpriseNumber.slice(0, 2) !== 'BE') return { isValid: false, error: 'Please fill in a Belgian enterprise number' };
// Remove country code
else enterpriseNumber = enterpriseNumber.substring(2);
}
//Check if first digit is 0
if (enterpriseNumber.length === 10 && enterpriseNumber.startsWith('0')) enterpriseNumber = enterpriseNumber.substring(1);
//Check if enterpriseNumber is valid with modulo test
if (parseInt(97 - (enterpriseNumber.slice(0, 7) % 97), 10) !== parseInt(enterpriseNumber.slice(7, 9), 10))
return { isValid: false, error: 'The provided number is invalid'}
return {
isValid: true,
enterpriseNumber: enterpriseNumber,
displayEnterpriseNumber: `BE 0${enterpriseNumber.substring(0, 3)}.${enterpriseNumber.substring(3, 6)}.${enterpriseNumber.substring(6, 9)}`
};
};
I think it's pretty messy and I'm wondering if this can be improved with one/two regex tests that reformat and validate the user's input?
A second question: Sometimes for account or credit cards numbers the input field had those underscores and lines (-) already in the input box and reformat the number while typing. What is this method called and can this be done for a specific thing like a Belgian enterprise number?
javascript regex reformatting
add a comment |
I want to reformat and validate if a user has provided a valid Belgian enterprise number. Because the input can be all of the following examples:
- BE 0123.321.123
- BE0123.321.123
- BE0123 321 123
- 0123.321.123
- 123.321.123
- 123321123
I've written a function that validates and reformat the input to a 'display' version (BE 0123.123.123) and a 'code' version (123123123). This function looks like this.
formatAndValidateEnterpriseNumber = enterpriseNumber => {
if(enterpriseNumber === undefined || !enterpriseNumber || (enterpriseNumber || '').length < 3) return { isValid: false, error: 'Please fill in your enterprise number' };
//Remove space, dots, ...
enterpriseNumber = enterpriseNumber.toUpperCase();
enterpriseNumber = enterpriseNumber.replace(/[. ,:-]+/g, '');
//Check for double country code
const reDouble = /^[a-zA-Z]{4}/;
if (reDouble.test(enterpriseNumber)) enterpriseNumber = enterpriseNumber.substring(2);
if (enterpriseNumber.length < 9 || enterpriseNumber.length > 12) return { isValid: false, error: 'The length of the provided number is incorrect' };
//Check country code
const reBE = /^[a-zA-Z]{2}/;
if (reBE.test(enterpriseNumber)) {
//Check if country code = BE
if (enterpriseNumber.slice(0, 2) !== 'BE') return { isValid: false, error: 'Please fill in a Belgian enterprise number' };
// Remove country code
else enterpriseNumber = enterpriseNumber.substring(2);
}
//Check if first digit is 0
if (enterpriseNumber.length === 10 && enterpriseNumber.startsWith('0')) enterpriseNumber = enterpriseNumber.substring(1);
//Check if enterpriseNumber is valid with modulo test
if (parseInt(97 - (enterpriseNumber.slice(0, 7) % 97), 10) !== parseInt(enterpriseNumber.slice(7, 9), 10))
return { isValid: false, error: 'The provided number is invalid'}
return {
isValid: true,
enterpriseNumber: enterpriseNumber,
displayEnterpriseNumber: `BE 0${enterpriseNumber.substring(0, 3)}.${enterpriseNumber.substring(3, 6)}.${enterpriseNumber.substring(6, 9)}`
};
};
I think it's pretty messy and I'm wondering if this can be improved with one/two regex tests that reformat and validate the user's input?
A second question: Sometimes for account or credit cards numbers the input field had those underscores and lines (-) already in the input box and reformat the number while typing. What is this method called and can this be done for a specific thing like a Belgian enterprise number?
javascript regex reformatting
The modulo test is not easily possible in regex.
– trincot
3 hours ago
add a comment |
I want to reformat and validate if a user has provided a valid Belgian enterprise number. Because the input can be all of the following examples:
- BE 0123.321.123
- BE0123.321.123
- BE0123 321 123
- 0123.321.123
- 123.321.123
- 123321123
I've written a function that validates and reformat the input to a 'display' version (BE 0123.123.123) and a 'code' version (123123123). This function looks like this.
formatAndValidateEnterpriseNumber = enterpriseNumber => {
if(enterpriseNumber === undefined || !enterpriseNumber || (enterpriseNumber || '').length < 3) return { isValid: false, error: 'Please fill in your enterprise number' };
//Remove space, dots, ...
enterpriseNumber = enterpriseNumber.toUpperCase();
enterpriseNumber = enterpriseNumber.replace(/[. ,:-]+/g, '');
//Check for double country code
const reDouble = /^[a-zA-Z]{4}/;
if (reDouble.test(enterpriseNumber)) enterpriseNumber = enterpriseNumber.substring(2);
if (enterpriseNumber.length < 9 || enterpriseNumber.length > 12) return { isValid: false, error: 'The length of the provided number is incorrect' };
//Check country code
const reBE = /^[a-zA-Z]{2}/;
if (reBE.test(enterpriseNumber)) {
//Check if country code = BE
if (enterpriseNumber.slice(0, 2) !== 'BE') return { isValid: false, error: 'Please fill in a Belgian enterprise number' };
// Remove country code
else enterpriseNumber = enterpriseNumber.substring(2);
}
//Check if first digit is 0
if (enterpriseNumber.length === 10 && enterpriseNumber.startsWith('0')) enterpriseNumber = enterpriseNumber.substring(1);
//Check if enterpriseNumber is valid with modulo test
if (parseInt(97 - (enterpriseNumber.slice(0, 7) % 97), 10) !== parseInt(enterpriseNumber.slice(7, 9), 10))
return { isValid: false, error: 'The provided number is invalid'}
return {
isValid: true,
enterpriseNumber: enterpriseNumber,
displayEnterpriseNumber: `BE 0${enterpriseNumber.substring(0, 3)}.${enterpriseNumber.substring(3, 6)}.${enterpriseNumber.substring(6, 9)}`
};
};
I think it's pretty messy and I'm wondering if this can be improved with one/two regex tests that reformat and validate the user's input?
A second question: Sometimes for account or credit cards numbers the input field had those underscores and lines (-) already in the input box and reformat the number while typing. What is this method called and can this be done for a specific thing like a Belgian enterprise number?
javascript regex reformatting
I want to reformat and validate if a user has provided a valid Belgian enterprise number. Because the input can be all of the following examples:
- BE 0123.321.123
- BE0123.321.123
- BE0123 321 123
- 0123.321.123
- 123.321.123
- 123321123
I've written a function that validates and reformat the input to a 'display' version (BE 0123.123.123) and a 'code' version (123123123). This function looks like this.
formatAndValidateEnterpriseNumber = enterpriseNumber => {
if(enterpriseNumber === undefined || !enterpriseNumber || (enterpriseNumber || '').length < 3) return { isValid: false, error: 'Please fill in your enterprise number' };
//Remove space, dots, ...
enterpriseNumber = enterpriseNumber.toUpperCase();
enterpriseNumber = enterpriseNumber.replace(/[. ,:-]+/g, '');
//Check for double country code
const reDouble = /^[a-zA-Z]{4}/;
if (reDouble.test(enterpriseNumber)) enterpriseNumber = enterpriseNumber.substring(2);
if (enterpriseNumber.length < 9 || enterpriseNumber.length > 12) return { isValid: false, error: 'The length of the provided number is incorrect' };
//Check country code
const reBE = /^[a-zA-Z]{2}/;
if (reBE.test(enterpriseNumber)) {
//Check if country code = BE
if (enterpriseNumber.slice(0, 2) !== 'BE') return { isValid: false, error: 'Please fill in a Belgian enterprise number' };
// Remove country code
else enterpriseNumber = enterpriseNumber.substring(2);
}
//Check if first digit is 0
if (enterpriseNumber.length === 10 && enterpriseNumber.startsWith('0')) enterpriseNumber = enterpriseNumber.substring(1);
//Check if enterpriseNumber is valid with modulo test
if (parseInt(97 - (enterpriseNumber.slice(0, 7) % 97), 10) !== parseInt(enterpriseNumber.slice(7, 9), 10))
return { isValid: false, error: 'The provided number is invalid'}
return {
isValid: true,
enterpriseNumber: enterpriseNumber,
displayEnterpriseNumber: `BE 0${enterpriseNumber.substring(0, 3)}.${enterpriseNumber.substring(3, 6)}.${enterpriseNumber.substring(6, 9)}`
};
};
I think it's pretty messy and I'm wondering if this can be improved with one/two regex tests that reformat and validate the user's input?
A second question: Sometimes for account or credit cards numbers the input field had those underscores and lines (-) already in the input box and reformat the number while typing. What is this method called and can this be done for a specific thing like a Belgian enterprise number?
javascript regex reformatting
javascript regex reformatting
asked 3 hours ago
ThoreThore
453213
453213
The modulo test is not easily possible in regex.
– trincot
3 hours ago
add a comment |
The modulo test is not easily possible in regex.
– trincot
3 hours ago
The modulo test is not easily possible in regex.
– trincot
3 hours ago
The modulo test is not easily possible in regex.
– trincot
3 hours ago
add a comment |
3 Answers
3
active
oldest
votes
Yes you can:
^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$
This regex should do it!
This source (in Dutch) states what an enterprise number is for Belgium:
It has the country code:
BEfollowed by a0or1and then followed by 9 digits.
https://regex101.com/r/4SRHxi/3
Explanation:
^and$: the string has to start and end with the given regex
(?:BE)?: look for a group with BE but?means it matches zero or
one times -?:means find but don't capture
s?: look for a space or dot that matches zero or one times
[0-1]?: check if a zero of one is present zero or one times
(d{3}): Check if succeeded by three digits
[.s]?: match a dot or a space zero or one times
(d{3}): Check if succeeded by three digits
[.s]?: again match a dot or a space zero or one times
(d{3}): check for last three digits
This will check if the input validates.
Editing it into the code version is simple:
«code».replace(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/g, "$1$2$3");
the g or global modifier will ensure all unwanted characters will be deleted. Here some magic happens. Since the three digits groups are surrounded by parenthesis the form a group. The (?:BE) does look for that group but doesn't capture it. We can put the groups back using the variables $1 to $3.
document.querySelector("pre").textContent.split("n").forEach(function(element){
if (element.match(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/))
{
console.log(element.replace(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/g, "$1$2$3"));
}
else
{
console.log(`REJECTED: ${element}`);
}
});<pre>
BE 0123.321.123
BE0123.321.123
BE0123 321 123
BE 01 23 32 11 23
BE 0123 32 11 23
1123.321.123
123.321.123
123321123
</pre>Rebuilding the String into the correct user friendly way is easy now:
document.querySelector("pre").textContent.split("n").forEach(function(element) {
if (element.match(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/)) {
var stripped = element.replace(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/g, "$1$2$3");
//with the modulo check from your code added back in.
if (97 - (parseInt(stripped.slice(0, 7), 10) % 97) == parseInt(stripped.slice(7, 9), 10)) {
//use a literal string
//use substring to put the dots between the sections of three numbers.
var humanReadable = `BE 0${stripped.substring(0,3)}.${stripped.substring(3,6)}.${stripped.substring(6,9)}`;
console.log(humanReadable);
}
}
});<pre>
BE 0123.321.123
BE0123.321.123
BE0123 321 123
0123.321.123
123.321.123
123321123
</pre>Second Question
Yes, this can be done however it requires you to write your own code for it.
Simple version:
document.querySelector("div.enterprisenumber > input").addEventListener("keydown", function(e) {
let value = this.value;
//prevent the input from going back to 0
if ( (value.length == 1 && (e.key == "Backspace" || e.key == "Delete"))) {
e.preventDefault();
return false;
}
}, true);
document.querySelector("div.enterprisenumber > input").addEventListener("keyup", function(e) {
//reset to a value without dots
let value = this.value.replace(/./g, "");
//strip the leading zero
const valueWithout = value.substring(1);
//calculate how much iterations we need of a groups of three digits.
const i = Math.floor(valueWithout.length / 3);
let newValue = "0";
//check if backspace or delete are used to make sure the dot can be deleted.
if (valueWithout.length < 10 && !(e.key == "Backspace" || e.key == "Delete")) {
//only fire when higher than zero
if (i > 0) {
let t;
//t is the index
for (t = 0; t < i; t++) {
//slice the correct portion of the string and append a dot, unless we are at the end of the groups
newValue += valueWithout.slice(t * 3, t * 3 + 3) + (t == 2 ? "" : ".");
}
//append the remainder that is not a group of three.
newValue += valueWithout.slice((t) * 3);
} else {
//return the value as is.
newValue = value;
}
//set the new value to the input.
this.value = newValue;
}
}, true);
document.querySelector("div.enterprisenumber > input").addEventListener("blur", function(e) {
let passed = false;
if (this.value.match(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/))
{
const value = this.value.substring(1).replace(/./g, "");
//with modulo check
if (97 - (parseInt(value.slice(0,7), 10) % 97) == value.slice(7, 9))
{
passed = true;
}
}
document.querySelector(".enterprisenumber").classList[(passed ? "remove" : "add")]("error");
});* {
box-sizing: border-box;
font-family: tahoma;
font-size: 10pt;
}
div.enterprisenumber {
border: 1px solid #747474;
width: 300px;
padding: 0px;
display: grid;
grid-template-columns: 25px auto;
}
div.enterprisenumber.error {
border: 1px solid #ff0000;
}
div.enterprisenumber>span {
grid-column: 1;
border: 0px;
padding: 3px;
background: linear-gradient(to right, rgba(0,0,0, 0.8) 33%, rgba(255,243,54, 0.8) 33%, rgba(255, 243, 54, 0.8) 66%, rgba(255, 15, 33, 0.8) 66%, rgba(255, 15, 33, 0.8) 100%);
color: #ffffff;
font-weight: bold;
text-shadow: 1px 1px #000000;
}
div.enterprisenumber>input {
grid-column: 2;
border: 0px;
padding: 3px;
}<div class="enterprisenumber">
<span>BE</span><input value="0" maxlength="12" />
</div>
Wow! Thanks for the explanation. I like the solution. However the modulo test is gone to check if the number is valid (The first 7 digits without the 0 or 1 in front, %97 should be the last two digits).
– Thore
1 hour ago
that's not that difficult, but not doable in the regexp though.
– Mouser
58 mins ago
Would you suggest to do it separately? First the regex and then another check for the validation of the numberparseInt(97 - (enterpriseNumber.slice(0, 7) % 97), 10) !== parseInt(enterpriseNumber.slice(7, 9), 10
– Thore
22 mins ago
1
See the solutions posted in my answer. Modulo check should be done separately.
– Mouser
21 mins ago
One last thing. I didn't come up with these examples in my code but something like BE 01 23 32 11 23 or a group separated with more than one space will return false?
– Thore
12 mins ago
|
show 1 more comment
For your example strings, you could match:
^(?:BEs?)?[01]?(d{3}([. ])d{3}2d{3}|d{9})$
That will match
^Start of string
(?:BEs?)?Optional BE followed by optional whitespace char
[01]?Optional zero or 1
(Capturing group
d{3}Match 3 digits
([. ])Capture in group either a space or digit to use as backreference
d{3}2d{3}Match 3 digits, dot or space (2 is the backreference) and 3 digits
|Or
d{9}Match 9 digits
)Close capturing group
$End of string
Regex demo
And then in the replacement use the first capturing group and replace the space or the dot with an empty string.
let pattern = /^(?:BEs?)?[01]?(d{3}([. ])d{3}2d{3}|d{9})$/;
let strings = [
"BE 0123.321.123",
"BE0123.321.123",
"BE0123 321 123",
"0123.321.123",
"123.321.123",
"123321123",
];
strings = strings.map(x => x.replace(pattern, function(m, g) {
let enterpriseNumber = g.replace(/[. ]/g, "");
return `BE 0${enterpriseNumber.substring(0, 3)}.${enterpriseNumber.substring(3, 6)}.${enterpriseNumber.substring(6, 9)}`
}));
console.log(strings);add a comment |
Here is an implementation of a BE ____.___.___ style of input. The pattern will be maintained, so the input will be guaranteed to have the "BE" prefix, the space, and the two dots. The validation can then concentrate on completeness and the modulo test.
Note that the input requires the first group to have 4 digits, where the first digit must be a 0 or a 1.
const ent = document.getElementById("ent");
const out = document.getElementById("isvalid");
function format() {
const re = /^D*[2-9]+|D+/g;
const [i, j] = [this.selectionStart, this.selectionEnd].map(i => {
i = this.value.slice(0, i).replace(re, "").length;
return i + 3 + (i >= 4 + format.backspace) + (i >= 7 + format.backspace);
});
this.value = "BE " + this.value.replace(re, "").padEnd(10, "_")
.replace(/(....)(...)(...).*/, "$1.$2.$3");
this.setSelectionRange(i, j);
format.backspace = false;
out.textContent = validate(this.value) ? "is valid" : "is invalid";
}
function validate(num) {
return /^BE [01](d{3}.){2}d{3}$/.test(num)
&& 97 - num.replace(/D/g, "").slice(0, 8) % 97 === +num.slice(-2);
}
ent.addEventListener("input", format);
ent.addEventListener("keydown", (e) => format.backspace = e.key == "Backspace");Belgian enterprise number: <input id="ent" value="BE ____.___.___">
<span id="isvalid"></span>
Wow, that's a nice solution to the second question.
– Mouser
20 mins ago
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Yes you can:
^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$
This regex should do it!
This source (in Dutch) states what an enterprise number is for Belgium:
It has the country code:
BEfollowed by a0or1and then followed by 9 digits.
https://regex101.com/r/4SRHxi/3
Explanation:
^and$: the string has to start and end with the given regex
(?:BE)?: look for a group with BE but?means it matches zero or
one times -?:means find but don't capture
s?: look for a space or dot that matches zero or one times
[0-1]?: check if a zero of one is present zero or one times
(d{3}): Check if succeeded by three digits
[.s]?: match a dot or a space zero or one times
(d{3}): Check if succeeded by three digits
[.s]?: again match a dot or a space zero or one times
(d{3}): check for last three digits
This will check if the input validates.
Editing it into the code version is simple:
«code».replace(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/g, "$1$2$3");
the g or global modifier will ensure all unwanted characters will be deleted. Here some magic happens. Since the three digits groups are surrounded by parenthesis the form a group. The (?:BE) does look for that group but doesn't capture it. We can put the groups back using the variables $1 to $3.
document.querySelector("pre").textContent.split("n").forEach(function(element){
if (element.match(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/))
{
console.log(element.replace(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/g, "$1$2$3"));
}
else
{
console.log(`REJECTED: ${element}`);
}
});<pre>
BE 0123.321.123
BE0123.321.123
BE0123 321 123
BE 01 23 32 11 23
BE 0123 32 11 23
1123.321.123
123.321.123
123321123
</pre>Rebuilding the String into the correct user friendly way is easy now:
document.querySelector("pre").textContent.split("n").forEach(function(element) {
if (element.match(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/)) {
var stripped = element.replace(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/g, "$1$2$3");
//with the modulo check from your code added back in.
if (97 - (parseInt(stripped.slice(0, 7), 10) % 97) == parseInt(stripped.slice(7, 9), 10)) {
//use a literal string
//use substring to put the dots between the sections of three numbers.
var humanReadable = `BE 0${stripped.substring(0,3)}.${stripped.substring(3,6)}.${stripped.substring(6,9)}`;
console.log(humanReadable);
}
}
});<pre>
BE 0123.321.123
BE0123.321.123
BE0123 321 123
0123.321.123
123.321.123
123321123
</pre>Second Question
Yes, this can be done however it requires you to write your own code for it.
Simple version:
document.querySelector("div.enterprisenumber > input").addEventListener("keydown", function(e) {
let value = this.value;
//prevent the input from going back to 0
if ( (value.length == 1 && (e.key == "Backspace" || e.key == "Delete"))) {
e.preventDefault();
return false;
}
}, true);
document.querySelector("div.enterprisenumber > input").addEventListener("keyup", function(e) {
//reset to a value without dots
let value = this.value.replace(/./g, "");
//strip the leading zero
const valueWithout = value.substring(1);
//calculate how much iterations we need of a groups of three digits.
const i = Math.floor(valueWithout.length / 3);
let newValue = "0";
//check if backspace or delete are used to make sure the dot can be deleted.
if (valueWithout.length < 10 && !(e.key == "Backspace" || e.key == "Delete")) {
//only fire when higher than zero
if (i > 0) {
let t;
//t is the index
for (t = 0; t < i; t++) {
//slice the correct portion of the string and append a dot, unless we are at the end of the groups
newValue += valueWithout.slice(t * 3, t * 3 + 3) + (t == 2 ? "" : ".");
}
//append the remainder that is not a group of three.
newValue += valueWithout.slice((t) * 3);
} else {
//return the value as is.
newValue = value;
}
//set the new value to the input.
this.value = newValue;
}
}, true);
document.querySelector("div.enterprisenumber > input").addEventListener("blur", function(e) {
let passed = false;
if (this.value.match(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/))
{
const value = this.value.substring(1).replace(/./g, "");
//with modulo check
if (97 - (parseInt(value.slice(0,7), 10) % 97) == value.slice(7, 9))
{
passed = true;
}
}
document.querySelector(".enterprisenumber").classList[(passed ? "remove" : "add")]("error");
});* {
box-sizing: border-box;
font-family: tahoma;
font-size: 10pt;
}
div.enterprisenumber {
border: 1px solid #747474;
width: 300px;
padding: 0px;
display: grid;
grid-template-columns: 25px auto;
}
div.enterprisenumber.error {
border: 1px solid #ff0000;
}
div.enterprisenumber>span {
grid-column: 1;
border: 0px;
padding: 3px;
background: linear-gradient(to right, rgba(0,0,0, 0.8) 33%, rgba(255,243,54, 0.8) 33%, rgba(255, 243, 54, 0.8) 66%, rgba(255, 15, 33, 0.8) 66%, rgba(255, 15, 33, 0.8) 100%);
color: #ffffff;
font-weight: bold;
text-shadow: 1px 1px #000000;
}
div.enterprisenumber>input {
grid-column: 2;
border: 0px;
padding: 3px;
}<div class="enterprisenumber">
<span>BE</span><input value="0" maxlength="12" />
</div>
Wow! Thanks for the explanation. I like the solution. However the modulo test is gone to check if the number is valid (The first 7 digits without the 0 or 1 in front, %97 should be the last two digits).
– Thore
1 hour ago
that's not that difficult, but not doable in the regexp though.
– Mouser
58 mins ago
Would you suggest to do it separately? First the regex and then another check for the validation of the numberparseInt(97 - (enterpriseNumber.slice(0, 7) % 97), 10) !== parseInt(enterpriseNumber.slice(7, 9), 10
– Thore
22 mins ago
1
See the solutions posted in my answer. Modulo check should be done separately.
– Mouser
21 mins ago
One last thing. I didn't come up with these examples in my code but something like BE 01 23 32 11 23 or a group separated with more than one space will return false?
– Thore
12 mins ago
|
show 1 more comment
Yes you can:
^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$
This regex should do it!
This source (in Dutch) states what an enterprise number is for Belgium:
It has the country code:
BEfollowed by a0or1and then followed by 9 digits.
https://regex101.com/r/4SRHxi/3
Explanation:
^and$: the string has to start and end with the given regex
(?:BE)?: look for a group with BE but?means it matches zero or
one times -?:means find but don't capture
s?: look for a space or dot that matches zero or one times
[0-1]?: check if a zero of one is present zero or one times
(d{3}): Check if succeeded by three digits
[.s]?: match a dot or a space zero or one times
(d{3}): Check if succeeded by three digits
[.s]?: again match a dot or a space zero or one times
(d{3}): check for last three digits
This will check if the input validates.
Editing it into the code version is simple:
«code».replace(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/g, "$1$2$3");
the g or global modifier will ensure all unwanted characters will be deleted. Here some magic happens. Since the three digits groups are surrounded by parenthesis the form a group. The (?:BE) does look for that group but doesn't capture it. We can put the groups back using the variables $1 to $3.
document.querySelector("pre").textContent.split("n").forEach(function(element){
if (element.match(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/))
{
console.log(element.replace(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/g, "$1$2$3"));
}
else
{
console.log(`REJECTED: ${element}`);
}
});<pre>
BE 0123.321.123
BE0123.321.123
BE0123 321 123
BE 01 23 32 11 23
BE 0123 32 11 23
1123.321.123
123.321.123
123321123
</pre>Rebuilding the String into the correct user friendly way is easy now:
document.querySelector("pre").textContent.split("n").forEach(function(element) {
if (element.match(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/)) {
var stripped = element.replace(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/g, "$1$2$3");
//with the modulo check from your code added back in.
if (97 - (parseInt(stripped.slice(0, 7), 10) % 97) == parseInt(stripped.slice(7, 9), 10)) {
//use a literal string
//use substring to put the dots between the sections of three numbers.
var humanReadable = `BE 0${stripped.substring(0,3)}.${stripped.substring(3,6)}.${stripped.substring(6,9)}`;
console.log(humanReadable);
}
}
});<pre>
BE 0123.321.123
BE0123.321.123
BE0123 321 123
0123.321.123
123.321.123
123321123
</pre>Second Question
Yes, this can be done however it requires you to write your own code for it.
Simple version:
document.querySelector("div.enterprisenumber > input").addEventListener("keydown", function(e) {
let value = this.value;
//prevent the input from going back to 0
if ( (value.length == 1 && (e.key == "Backspace" || e.key == "Delete"))) {
e.preventDefault();
return false;
}
}, true);
document.querySelector("div.enterprisenumber > input").addEventListener("keyup", function(e) {
//reset to a value without dots
let value = this.value.replace(/./g, "");
//strip the leading zero
const valueWithout = value.substring(1);
//calculate how much iterations we need of a groups of three digits.
const i = Math.floor(valueWithout.length / 3);
let newValue = "0";
//check if backspace or delete are used to make sure the dot can be deleted.
if (valueWithout.length < 10 && !(e.key == "Backspace" || e.key == "Delete")) {
//only fire when higher than zero
if (i > 0) {
let t;
//t is the index
for (t = 0; t < i; t++) {
//slice the correct portion of the string and append a dot, unless we are at the end of the groups
newValue += valueWithout.slice(t * 3, t * 3 + 3) + (t == 2 ? "" : ".");
}
//append the remainder that is not a group of three.
newValue += valueWithout.slice((t) * 3);
} else {
//return the value as is.
newValue = value;
}
//set the new value to the input.
this.value = newValue;
}
}, true);
document.querySelector("div.enterprisenumber > input").addEventListener("blur", function(e) {
let passed = false;
if (this.value.match(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/))
{
const value = this.value.substring(1).replace(/./g, "");
//with modulo check
if (97 - (parseInt(value.slice(0,7), 10) % 97) == value.slice(7, 9))
{
passed = true;
}
}
document.querySelector(".enterprisenumber").classList[(passed ? "remove" : "add")]("error");
});* {
box-sizing: border-box;
font-family: tahoma;
font-size: 10pt;
}
div.enterprisenumber {
border: 1px solid #747474;
width: 300px;
padding: 0px;
display: grid;
grid-template-columns: 25px auto;
}
div.enterprisenumber.error {
border: 1px solid #ff0000;
}
div.enterprisenumber>span {
grid-column: 1;
border: 0px;
padding: 3px;
background: linear-gradient(to right, rgba(0,0,0, 0.8) 33%, rgba(255,243,54, 0.8) 33%, rgba(255, 243, 54, 0.8) 66%, rgba(255, 15, 33, 0.8) 66%, rgba(255, 15, 33, 0.8) 100%);
color: #ffffff;
font-weight: bold;
text-shadow: 1px 1px #000000;
}
div.enterprisenumber>input {
grid-column: 2;
border: 0px;
padding: 3px;
}<div class="enterprisenumber">
<span>BE</span><input value="0" maxlength="12" />
</div>
Wow! Thanks for the explanation. I like the solution. However the modulo test is gone to check if the number is valid (The first 7 digits without the 0 or 1 in front, %97 should be the last two digits).
– Thore
1 hour ago
that's not that difficult, but not doable in the regexp though.
– Mouser
58 mins ago
Would you suggest to do it separately? First the regex and then another check for the validation of the numberparseInt(97 - (enterpriseNumber.slice(0, 7) % 97), 10) !== parseInt(enterpriseNumber.slice(7, 9), 10
– Thore
22 mins ago
1
See the solutions posted in my answer. Modulo check should be done separately.
– Mouser
21 mins ago
One last thing. I didn't come up with these examples in my code but something like BE 01 23 32 11 23 or a group separated with more than one space will return false?
– Thore
12 mins ago
|
show 1 more comment
Yes you can:
^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$
This regex should do it!
This source (in Dutch) states what an enterprise number is for Belgium:
It has the country code:
BEfollowed by a0or1and then followed by 9 digits.
https://regex101.com/r/4SRHxi/3
Explanation:
^and$: the string has to start and end with the given regex
(?:BE)?: look for a group with BE but?means it matches zero or
one times -?:means find but don't capture
s?: look for a space or dot that matches zero or one times
[0-1]?: check if a zero of one is present zero or one times
(d{3}): Check if succeeded by three digits
[.s]?: match a dot or a space zero or one times
(d{3}): Check if succeeded by three digits
[.s]?: again match a dot or a space zero or one times
(d{3}): check for last three digits
This will check if the input validates.
Editing it into the code version is simple:
«code».replace(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/g, "$1$2$3");
the g or global modifier will ensure all unwanted characters will be deleted. Here some magic happens. Since the three digits groups are surrounded by parenthesis the form a group. The (?:BE) does look for that group but doesn't capture it. We can put the groups back using the variables $1 to $3.
document.querySelector("pre").textContent.split("n").forEach(function(element){
if (element.match(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/))
{
console.log(element.replace(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/g, "$1$2$3"));
}
else
{
console.log(`REJECTED: ${element}`);
}
});<pre>
BE 0123.321.123
BE0123.321.123
BE0123 321 123
BE 01 23 32 11 23
BE 0123 32 11 23
1123.321.123
123.321.123
123321123
</pre>Rebuilding the String into the correct user friendly way is easy now:
document.querySelector("pre").textContent.split("n").forEach(function(element) {
if (element.match(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/)) {
var stripped = element.replace(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/g, "$1$2$3");
//with the modulo check from your code added back in.
if (97 - (parseInt(stripped.slice(0, 7), 10) % 97) == parseInt(stripped.slice(7, 9), 10)) {
//use a literal string
//use substring to put the dots between the sections of three numbers.
var humanReadable = `BE 0${stripped.substring(0,3)}.${stripped.substring(3,6)}.${stripped.substring(6,9)}`;
console.log(humanReadable);
}
}
});<pre>
BE 0123.321.123
BE0123.321.123
BE0123 321 123
0123.321.123
123.321.123
123321123
</pre>Second Question
Yes, this can be done however it requires you to write your own code for it.
Simple version:
document.querySelector("div.enterprisenumber > input").addEventListener("keydown", function(e) {
let value = this.value;
//prevent the input from going back to 0
if ( (value.length == 1 && (e.key == "Backspace" || e.key == "Delete"))) {
e.preventDefault();
return false;
}
}, true);
document.querySelector("div.enterprisenumber > input").addEventListener("keyup", function(e) {
//reset to a value without dots
let value = this.value.replace(/./g, "");
//strip the leading zero
const valueWithout = value.substring(1);
//calculate how much iterations we need of a groups of three digits.
const i = Math.floor(valueWithout.length / 3);
let newValue = "0";
//check if backspace or delete are used to make sure the dot can be deleted.
if (valueWithout.length < 10 && !(e.key == "Backspace" || e.key == "Delete")) {
//only fire when higher than zero
if (i > 0) {
let t;
//t is the index
for (t = 0; t < i; t++) {
//slice the correct portion of the string and append a dot, unless we are at the end of the groups
newValue += valueWithout.slice(t * 3, t * 3 + 3) + (t == 2 ? "" : ".");
}
//append the remainder that is not a group of three.
newValue += valueWithout.slice((t) * 3);
} else {
//return the value as is.
newValue = value;
}
//set the new value to the input.
this.value = newValue;
}
}, true);
document.querySelector("div.enterprisenumber > input").addEventListener("blur", function(e) {
let passed = false;
if (this.value.match(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/))
{
const value = this.value.substring(1).replace(/./g, "");
//with modulo check
if (97 - (parseInt(value.slice(0,7), 10) % 97) == value.slice(7, 9))
{
passed = true;
}
}
document.querySelector(".enterprisenumber").classList[(passed ? "remove" : "add")]("error");
});* {
box-sizing: border-box;
font-family: tahoma;
font-size: 10pt;
}
div.enterprisenumber {
border: 1px solid #747474;
width: 300px;
padding: 0px;
display: grid;
grid-template-columns: 25px auto;
}
div.enterprisenumber.error {
border: 1px solid #ff0000;
}
div.enterprisenumber>span {
grid-column: 1;
border: 0px;
padding: 3px;
background: linear-gradient(to right, rgba(0,0,0, 0.8) 33%, rgba(255,243,54, 0.8) 33%, rgba(255, 243, 54, 0.8) 66%, rgba(255, 15, 33, 0.8) 66%, rgba(255, 15, 33, 0.8) 100%);
color: #ffffff;
font-weight: bold;
text-shadow: 1px 1px #000000;
}
div.enterprisenumber>input {
grid-column: 2;
border: 0px;
padding: 3px;
}<div class="enterprisenumber">
<span>BE</span><input value="0" maxlength="12" />
</div>Yes you can:
^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$
This regex should do it!
This source (in Dutch) states what an enterprise number is for Belgium:
It has the country code:
BEfollowed by a0or1and then followed by 9 digits.
https://regex101.com/r/4SRHxi/3
Explanation:
^and$: the string has to start and end with the given regex
(?:BE)?: look for a group with BE but?means it matches zero or
one times -?:means find but don't capture
s?: look for a space or dot that matches zero or one times
[0-1]?: check if a zero of one is present zero or one times
(d{3}): Check if succeeded by three digits
[.s]?: match a dot or a space zero or one times
(d{3}): Check if succeeded by three digits
[.s]?: again match a dot or a space zero or one times
(d{3}): check for last three digits
This will check if the input validates.
Editing it into the code version is simple:
«code».replace(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/g, "$1$2$3");
the g or global modifier will ensure all unwanted characters will be deleted. Here some magic happens. Since the three digits groups are surrounded by parenthesis the form a group. The (?:BE) does look for that group but doesn't capture it. We can put the groups back using the variables $1 to $3.
document.querySelector("pre").textContent.split("n").forEach(function(element){
if (element.match(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/))
{
console.log(element.replace(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/g, "$1$2$3"));
}
else
{
console.log(`REJECTED: ${element}`);
}
});<pre>
BE 0123.321.123
BE0123.321.123
BE0123 321 123
BE 01 23 32 11 23
BE 0123 32 11 23
1123.321.123
123.321.123
123321123
</pre>Rebuilding the String into the correct user friendly way is easy now:
document.querySelector("pre").textContent.split("n").forEach(function(element) {
if (element.match(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/)) {
var stripped = element.replace(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/g, "$1$2$3");
//with the modulo check from your code added back in.
if (97 - (parseInt(stripped.slice(0, 7), 10) % 97) == parseInt(stripped.slice(7, 9), 10)) {
//use a literal string
//use substring to put the dots between the sections of three numbers.
var humanReadable = `BE 0${stripped.substring(0,3)}.${stripped.substring(3,6)}.${stripped.substring(6,9)}`;
console.log(humanReadable);
}
}
});<pre>
BE 0123.321.123
BE0123.321.123
BE0123 321 123
0123.321.123
123.321.123
123321123
</pre>Second Question
Yes, this can be done however it requires you to write your own code for it.
Simple version:
document.querySelector("div.enterprisenumber > input").addEventListener("keydown", function(e) {
let value = this.value;
//prevent the input from going back to 0
if ( (value.length == 1 && (e.key == "Backspace" || e.key == "Delete"))) {
e.preventDefault();
return false;
}
}, true);
document.querySelector("div.enterprisenumber > input").addEventListener("keyup", function(e) {
//reset to a value without dots
let value = this.value.replace(/./g, "");
//strip the leading zero
const valueWithout = value.substring(1);
//calculate how much iterations we need of a groups of three digits.
const i = Math.floor(valueWithout.length / 3);
let newValue = "0";
//check if backspace or delete are used to make sure the dot can be deleted.
if (valueWithout.length < 10 && !(e.key == "Backspace" || e.key == "Delete")) {
//only fire when higher than zero
if (i > 0) {
let t;
//t is the index
for (t = 0; t < i; t++) {
//slice the correct portion of the string and append a dot, unless we are at the end of the groups
newValue += valueWithout.slice(t * 3, t * 3 + 3) + (t == 2 ? "" : ".");
}
//append the remainder that is not a group of three.
newValue += valueWithout.slice((t) * 3);
} else {
//return the value as is.
newValue = value;
}
//set the new value to the input.
this.value = newValue;
}
}, true);
document.querySelector("div.enterprisenumber > input").addEventListener("blur", function(e) {
let passed = false;
if (this.value.match(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/))
{
const value = this.value.substring(1).replace(/./g, "");
//with modulo check
if (97 - (parseInt(value.slice(0,7), 10) % 97) == value.slice(7, 9))
{
passed = true;
}
}
document.querySelector(".enterprisenumber").classList[(passed ? "remove" : "add")]("error");
});* {
box-sizing: border-box;
font-family: tahoma;
font-size: 10pt;
}
div.enterprisenumber {
border: 1px solid #747474;
width: 300px;
padding: 0px;
display: grid;
grid-template-columns: 25px auto;
}
div.enterprisenumber.error {
border: 1px solid #ff0000;
}
div.enterprisenumber>span {
grid-column: 1;
border: 0px;
padding: 3px;
background: linear-gradient(to right, rgba(0,0,0, 0.8) 33%, rgba(255,243,54, 0.8) 33%, rgba(255, 243, 54, 0.8) 66%, rgba(255, 15, 33, 0.8) 66%, rgba(255, 15, 33, 0.8) 100%);
color: #ffffff;
font-weight: bold;
text-shadow: 1px 1px #000000;
}
div.enterprisenumber>input {
grid-column: 2;
border: 0px;
padding: 3px;
}<div class="enterprisenumber">
<span>BE</span><input value="0" maxlength="12" />
</div>document.querySelector("pre").textContent.split("n").forEach(function(element){
if (element.match(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/))
{
console.log(element.replace(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/g, "$1$2$3"));
}
else
{
console.log(`REJECTED: ${element}`);
}
});<pre>
BE 0123.321.123
BE0123.321.123
BE0123 321 123
BE 01 23 32 11 23
BE 0123 32 11 23
1123.321.123
123.321.123
123321123
</pre>document.querySelector("pre").textContent.split("n").forEach(function(element){
if (element.match(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/))
{
console.log(element.replace(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/g, "$1$2$3"));
}
else
{
console.log(`REJECTED: ${element}`);
}
});<pre>
BE 0123.321.123
BE0123.321.123
BE0123 321 123
BE 01 23 32 11 23
BE 0123 32 11 23
1123.321.123
123.321.123
123321123
</pre>document.querySelector("pre").textContent.split("n").forEach(function(element) {
if (element.match(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/)) {
var stripped = element.replace(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/g, "$1$2$3");
//with the modulo check from your code added back in.
if (97 - (parseInt(stripped.slice(0, 7), 10) % 97) == parseInt(stripped.slice(7, 9), 10)) {
//use a literal string
//use substring to put the dots between the sections of three numbers.
var humanReadable = `BE 0${stripped.substring(0,3)}.${stripped.substring(3,6)}.${stripped.substring(6,9)}`;
console.log(humanReadable);
}
}
});<pre>
BE 0123.321.123
BE0123.321.123
BE0123 321 123
0123.321.123
123.321.123
123321123
</pre>document.querySelector("pre").textContent.split("n").forEach(function(element) {
if (element.match(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/)) {
var stripped = element.replace(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/g, "$1$2$3");
//with the modulo check from your code added back in.
if (97 - (parseInt(stripped.slice(0, 7), 10) % 97) == parseInt(stripped.slice(7, 9), 10)) {
//use a literal string
//use substring to put the dots between the sections of three numbers.
var humanReadable = `BE 0${stripped.substring(0,3)}.${stripped.substring(3,6)}.${stripped.substring(6,9)}`;
console.log(humanReadable);
}
}
});<pre>
BE 0123.321.123
BE0123.321.123
BE0123 321 123
0123.321.123
123.321.123
123321123
</pre>document.querySelector("div.enterprisenumber > input").addEventListener("keydown", function(e) {
let value = this.value;
//prevent the input from going back to 0
if ( (value.length == 1 && (e.key == "Backspace" || e.key == "Delete"))) {
e.preventDefault();
return false;
}
}, true);
document.querySelector("div.enterprisenumber > input").addEventListener("keyup", function(e) {
//reset to a value without dots
let value = this.value.replace(/./g, "");
//strip the leading zero
const valueWithout = value.substring(1);
//calculate how much iterations we need of a groups of three digits.
const i = Math.floor(valueWithout.length / 3);
let newValue = "0";
//check if backspace or delete are used to make sure the dot can be deleted.
if (valueWithout.length < 10 && !(e.key == "Backspace" || e.key == "Delete")) {
//only fire when higher than zero
if (i > 0) {
let t;
//t is the index
for (t = 0; t < i; t++) {
//slice the correct portion of the string and append a dot, unless we are at the end of the groups
newValue += valueWithout.slice(t * 3, t * 3 + 3) + (t == 2 ? "" : ".");
}
//append the remainder that is not a group of three.
newValue += valueWithout.slice((t) * 3);
} else {
//return the value as is.
newValue = value;
}
//set the new value to the input.
this.value = newValue;
}
}, true);
document.querySelector("div.enterprisenumber > input").addEventListener("blur", function(e) {
let passed = false;
if (this.value.match(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/))
{
const value = this.value.substring(1).replace(/./g, "");
//with modulo check
if (97 - (parseInt(value.slice(0,7), 10) % 97) == value.slice(7, 9))
{
passed = true;
}
}
document.querySelector(".enterprisenumber").classList[(passed ? "remove" : "add")]("error");
});* {
box-sizing: border-box;
font-family: tahoma;
font-size: 10pt;
}
div.enterprisenumber {
border: 1px solid #747474;
width: 300px;
padding: 0px;
display: grid;
grid-template-columns: 25px auto;
}
div.enterprisenumber.error {
border: 1px solid #ff0000;
}
div.enterprisenumber>span {
grid-column: 1;
border: 0px;
padding: 3px;
background: linear-gradient(to right, rgba(0,0,0, 0.8) 33%, rgba(255,243,54, 0.8) 33%, rgba(255, 243, 54, 0.8) 66%, rgba(255, 15, 33, 0.8) 66%, rgba(255, 15, 33, 0.8) 100%);
color: #ffffff;
font-weight: bold;
text-shadow: 1px 1px #000000;
}
div.enterprisenumber>input {
grid-column: 2;
border: 0px;
padding: 3px;
}<div class="enterprisenumber">
<span>BE</span><input value="0" maxlength="12" />
</div>document.querySelector("div.enterprisenumber > input").addEventListener("keydown", function(e) {
let value = this.value;
//prevent the input from going back to 0
if ( (value.length == 1 && (e.key == "Backspace" || e.key == "Delete"))) {
e.preventDefault();
return false;
}
}, true);
document.querySelector("div.enterprisenumber > input").addEventListener("keyup", function(e) {
//reset to a value without dots
let value = this.value.replace(/./g, "");
//strip the leading zero
const valueWithout = value.substring(1);
//calculate how much iterations we need of a groups of three digits.
const i = Math.floor(valueWithout.length / 3);
let newValue = "0";
//check if backspace or delete are used to make sure the dot can be deleted.
if (valueWithout.length < 10 && !(e.key == "Backspace" || e.key == "Delete")) {
//only fire when higher than zero
if (i > 0) {
let t;
//t is the index
for (t = 0; t < i; t++) {
//slice the correct portion of the string and append a dot, unless we are at the end of the groups
newValue += valueWithout.slice(t * 3, t * 3 + 3) + (t == 2 ? "" : ".");
}
//append the remainder that is not a group of three.
newValue += valueWithout.slice((t) * 3);
} else {
//return the value as is.
newValue = value;
}
//set the new value to the input.
this.value = newValue;
}
}, true);
document.querySelector("div.enterprisenumber > input").addEventListener("blur", function(e) {
let passed = false;
if (this.value.match(/^(?:BE)?s?[0-1]?(d{3})[.s]?(d{3})[.s]?(d{3})$/))
{
const value = this.value.substring(1).replace(/./g, "");
//with modulo check
if (97 - (parseInt(value.slice(0,7), 10) % 97) == value.slice(7, 9))
{
passed = true;
}
}
document.querySelector(".enterprisenumber").classList[(passed ? "remove" : "add")]("error");
});* {
box-sizing: border-box;
font-family: tahoma;
font-size: 10pt;
}
div.enterprisenumber {
border: 1px solid #747474;
width: 300px;
padding: 0px;
display: grid;
grid-template-columns: 25px auto;
}
div.enterprisenumber.error {
border: 1px solid #ff0000;
}
div.enterprisenumber>span {
grid-column: 1;
border: 0px;
padding: 3px;
background: linear-gradient(to right, rgba(0,0,0, 0.8) 33%, rgba(255,243,54, 0.8) 33%, rgba(255, 243, 54, 0.8) 66%, rgba(255, 15, 33, 0.8) 66%, rgba(255, 15, 33, 0.8) 100%);
color: #ffffff;
font-weight: bold;
text-shadow: 1px 1px #000000;
}
div.enterprisenumber>input {
grid-column: 2;
border: 0px;
padding: 3px;
}<div class="enterprisenumber">
<span>BE</span><input value="0" maxlength="12" />
</div>edited 2 mins ago
answered 3 hours ago
MouserMouser
11.5k21948
11.5k21948
Wow! Thanks for the explanation. I like the solution. However the modulo test is gone to check if the number is valid (The first 7 digits without the 0 or 1 in front, %97 should be the last two digits).
– Thore
1 hour ago
that's not that difficult, but not doable in the regexp though.
– Mouser
58 mins ago
Would you suggest to do it separately? First the regex and then another check for the validation of the numberparseInt(97 - (enterpriseNumber.slice(0, 7) % 97), 10) !== parseInt(enterpriseNumber.slice(7, 9), 10
– Thore
22 mins ago
1
See the solutions posted in my answer. Modulo check should be done separately.
– Mouser
21 mins ago
One last thing. I didn't come up with these examples in my code but something like BE 01 23 32 11 23 or a group separated with more than one space will return false?
– Thore
12 mins ago
|
show 1 more comment
Wow! Thanks for the explanation. I like the solution. However the modulo test is gone to check if the number is valid (The first 7 digits without the 0 or 1 in front, %97 should be the last two digits).
– Thore
1 hour ago
that's not that difficult, but not doable in the regexp though.
– Mouser
58 mins ago
Would you suggest to do it separately? First the regex and then another check for the validation of the numberparseInt(97 - (enterpriseNumber.slice(0, 7) % 97), 10) !== parseInt(enterpriseNumber.slice(7, 9), 10
– Thore
22 mins ago
1
See the solutions posted in my answer. Modulo check should be done separately.
– Mouser
21 mins ago
One last thing. I didn't come up with these examples in my code but something like BE 01 23 32 11 23 or a group separated with more than one space will return false?
– Thore
12 mins ago
Wow! Thanks for the explanation. I like the solution. However the modulo test is gone to check if the number is valid (The first 7 digits without the 0 or 1 in front, %97 should be the last two digits).
– Thore
1 hour ago
Wow! Thanks for the explanation. I like the solution. However the modulo test is gone to check if the number is valid (The first 7 digits without the 0 or 1 in front, %97 should be the last two digits).
– Thore
1 hour ago
that's not that difficult, but not doable in the regexp though.
– Mouser
58 mins ago
that's not that difficult, but not doable in the regexp though.
– Mouser
58 mins ago
Would you suggest to do it separately? First the regex and then another check for the validation of the number
parseInt(97 - (enterpriseNumber.slice(0, 7) % 97), 10) !== parseInt(enterpriseNumber.slice(7, 9), 10– Thore
22 mins ago
Would you suggest to do it separately? First the regex and then another check for the validation of the number
parseInt(97 - (enterpriseNumber.slice(0, 7) % 97), 10) !== parseInt(enterpriseNumber.slice(7, 9), 10– Thore
22 mins ago
1
1
See the solutions posted in my answer. Modulo check should be done separately.
– Mouser
21 mins ago
See the solutions posted in my answer. Modulo check should be done separately.
– Mouser
21 mins ago
One last thing. I didn't come up with these examples in my code but something like BE 01 23 32 11 23 or a group separated with more than one space will return false?
– Thore
12 mins ago
One last thing. I didn't come up with these examples in my code but something like BE 01 23 32 11 23 or a group separated with more than one space will return false?
– Thore
12 mins ago
|
show 1 more comment
For your example strings, you could match:
^(?:BEs?)?[01]?(d{3}([. ])d{3}2d{3}|d{9})$
That will match
^Start of string
(?:BEs?)?Optional BE followed by optional whitespace char
[01]?Optional zero or 1
(Capturing group
d{3}Match 3 digits
([. ])Capture in group either a space or digit to use as backreference
d{3}2d{3}Match 3 digits, dot or space (2 is the backreference) and 3 digits
|Or
d{9}Match 9 digits
)Close capturing group
$End of string
Regex demo
And then in the replacement use the first capturing group and replace the space or the dot with an empty string.
let pattern = /^(?:BEs?)?[01]?(d{3}([. ])d{3}2d{3}|d{9})$/;
let strings = [
"BE 0123.321.123",
"BE0123.321.123",
"BE0123 321 123",
"0123.321.123",
"123.321.123",
"123321123",
];
strings = strings.map(x => x.replace(pattern, function(m, g) {
let enterpriseNumber = g.replace(/[. ]/g, "");
return `BE 0${enterpriseNumber.substring(0, 3)}.${enterpriseNumber.substring(3, 6)}.${enterpriseNumber.substring(6, 9)}`
}));
console.log(strings);add a comment |
For your example strings, you could match:
^(?:BEs?)?[01]?(d{3}([. ])d{3}2d{3}|d{9})$
That will match
^Start of string
(?:BEs?)?Optional BE followed by optional whitespace char
[01]?Optional zero or 1
(Capturing group
d{3}Match 3 digits
([. ])Capture in group either a space or digit to use as backreference
d{3}2d{3}Match 3 digits, dot or space (2 is the backreference) and 3 digits
|Or
d{9}Match 9 digits
)Close capturing group
$End of string
Regex demo
And then in the replacement use the first capturing group and replace the space or the dot with an empty string.
let pattern = /^(?:BEs?)?[01]?(d{3}([. ])d{3}2d{3}|d{9})$/;
let strings = [
"BE 0123.321.123",
"BE0123.321.123",
"BE0123 321 123",
"0123.321.123",
"123.321.123",
"123321123",
];
strings = strings.map(x => x.replace(pattern, function(m, g) {
let enterpriseNumber = g.replace(/[. ]/g, "");
return `BE 0${enterpriseNumber.substring(0, 3)}.${enterpriseNumber.substring(3, 6)}.${enterpriseNumber.substring(6, 9)}`
}));
console.log(strings);add a comment |
For your example strings, you could match:
^(?:BEs?)?[01]?(d{3}([. ])d{3}2d{3}|d{9})$
That will match
^Start of string
(?:BEs?)?Optional BE followed by optional whitespace char
[01]?Optional zero or 1
(Capturing group
d{3}Match 3 digits
([. ])Capture in group either a space or digit to use as backreference
d{3}2d{3}Match 3 digits, dot or space (2 is the backreference) and 3 digits
|Or
d{9}Match 9 digits
)Close capturing group
$End of string
Regex demo
And then in the replacement use the first capturing group and replace the space or the dot with an empty string.
let pattern = /^(?:BEs?)?[01]?(d{3}([. ])d{3}2d{3}|d{9})$/;
let strings = [
"BE 0123.321.123",
"BE0123.321.123",
"BE0123 321 123",
"0123.321.123",
"123.321.123",
"123321123",
];
strings = strings.map(x => x.replace(pattern, function(m, g) {
let enterpriseNumber = g.replace(/[. ]/g, "");
return `BE 0${enterpriseNumber.substring(0, 3)}.${enterpriseNumber.substring(3, 6)}.${enterpriseNumber.substring(6, 9)}`
}));
console.log(strings);For your example strings, you could match:
^(?:BEs?)?[01]?(d{3}([. ])d{3}2d{3}|d{9})$
That will match
^Start of string
(?:BEs?)?Optional BE followed by optional whitespace char
[01]?Optional zero or 1
(Capturing group
d{3}Match 3 digits
([. ])Capture in group either a space or digit to use as backreference
d{3}2d{3}Match 3 digits, dot or space (2 is the backreference) and 3 digits
|Or
d{9}Match 9 digits
)Close capturing group
$End of string
Regex demo
And then in the replacement use the first capturing group and replace the space or the dot with an empty string.
let pattern = /^(?:BEs?)?[01]?(d{3}([. ])d{3}2d{3}|d{9})$/;
let strings = [
"BE 0123.321.123",
"BE0123.321.123",
"BE0123 321 123",
"0123.321.123",
"123.321.123",
"123321123",
];
strings = strings.map(x => x.replace(pattern, function(m, g) {
let enterpriseNumber = g.replace(/[. ]/g, "");
return `BE 0${enterpriseNumber.substring(0, 3)}.${enterpriseNumber.substring(3, 6)}.${enterpriseNumber.substring(6, 9)}`
}));
console.log(strings);let pattern = /^(?:BEs?)?[01]?(d{3}([. ])d{3}2d{3}|d{9})$/;
let strings = [
"BE 0123.321.123",
"BE0123.321.123",
"BE0123 321 123",
"0123.321.123",
"123.321.123",
"123321123",
];
strings = strings.map(x => x.replace(pattern, function(m, g) {
let enterpriseNumber = g.replace(/[. ]/g, "");
return `BE 0${enterpriseNumber.substring(0, 3)}.${enterpriseNumber.substring(3, 6)}.${enterpriseNumber.substring(6, 9)}`
}));
console.log(strings);let pattern = /^(?:BEs?)?[01]?(d{3}([. ])d{3}2d{3}|d{9})$/;
let strings = [
"BE 0123.321.123",
"BE0123.321.123",
"BE0123 321 123",
"0123.321.123",
"123.321.123",
"123321123",
];
strings = strings.map(x => x.replace(pattern, function(m, g) {
let enterpriseNumber = g.replace(/[. ]/g, "");
return `BE 0${enterpriseNumber.substring(0, 3)}.${enterpriseNumber.substring(3, 6)}.${enterpriseNumber.substring(6, 9)}`
}));
console.log(strings);edited 1 hour ago
answered 3 hours ago
The fourth birdThe fourth bird
23.8k81429
23.8k81429
add a comment |
add a comment |
Here is an implementation of a BE ____.___.___ style of input. The pattern will be maintained, so the input will be guaranteed to have the "BE" prefix, the space, and the two dots. The validation can then concentrate on completeness and the modulo test.
Note that the input requires the first group to have 4 digits, where the first digit must be a 0 or a 1.
const ent = document.getElementById("ent");
const out = document.getElementById("isvalid");
function format() {
const re = /^D*[2-9]+|D+/g;
const [i, j] = [this.selectionStart, this.selectionEnd].map(i => {
i = this.value.slice(0, i).replace(re, "").length;
return i + 3 + (i >= 4 + format.backspace) + (i >= 7 + format.backspace);
});
this.value = "BE " + this.value.replace(re, "").padEnd(10, "_")
.replace(/(....)(...)(...).*/, "$1.$2.$3");
this.setSelectionRange(i, j);
format.backspace = false;
out.textContent = validate(this.value) ? "is valid" : "is invalid";
}
function validate(num) {
return /^BE [01](d{3}.){2}d{3}$/.test(num)
&& 97 - num.replace(/D/g, "").slice(0, 8) % 97 === +num.slice(-2);
}
ent.addEventListener("input", format);
ent.addEventListener("keydown", (e) => format.backspace = e.key == "Backspace");Belgian enterprise number: <input id="ent" value="BE ____.___.___">
<span id="isvalid"></span>
Wow, that's a nice solution to the second question.
– Mouser
20 mins ago
add a comment |
Here is an implementation of a BE ____.___.___ style of input. The pattern will be maintained, so the input will be guaranteed to have the "BE" prefix, the space, and the two dots. The validation can then concentrate on completeness and the modulo test.
Note that the input requires the first group to have 4 digits, where the first digit must be a 0 or a 1.
const ent = document.getElementById("ent");
const out = document.getElementById("isvalid");
function format() {
const re = /^D*[2-9]+|D+/g;
const [i, j] = [this.selectionStart, this.selectionEnd].map(i => {
i = this.value.slice(0, i).replace(re, "").length;
return i + 3 + (i >= 4 + format.backspace) + (i >= 7 + format.backspace);
});
this.value = "BE " + this.value.replace(re, "").padEnd(10, "_")
.replace(/(....)(...)(...).*/, "$1.$2.$3");
this.setSelectionRange(i, j);
format.backspace = false;
out.textContent = validate(this.value) ? "is valid" : "is invalid";
}
function validate(num) {
return /^BE [01](d{3}.){2}d{3}$/.test(num)
&& 97 - num.replace(/D/g, "").slice(0, 8) % 97 === +num.slice(-2);
}
ent.addEventListener("input", format);
ent.addEventListener("keydown", (e) => format.backspace = e.key == "Backspace");Belgian enterprise number: <input id="ent" value="BE ____.___.___">
<span id="isvalid"></span>
Wow, that's a nice solution to the second question.
– Mouser
20 mins ago
add a comment |
Here is an implementation of a BE ____.___.___ style of input. The pattern will be maintained, so the input will be guaranteed to have the "BE" prefix, the space, and the two dots. The validation can then concentrate on completeness and the modulo test.
Note that the input requires the first group to have 4 digits, where the first digit must be a 0 or a 1.
const ent = document.getElementById("ent");
const out = document.getElementById("isvalid");
function format() {
const re = /^D*[2-9]+|D+/g;
const [i, j] = [this.selectionStart, this.selectionEnd].map(i => {
i = this.value.slice(0, i).replace(re, "").length;
return i + 3 + (i >= 4 + format.backspace) + (i >= 7 + format.backspace);
});
this.value = "BE " + this.value.replace(re, "").padEnd(10, "_")
.replace(/(....)(...)(...).*/, "$1.$2.$3");
this.setSelectionRange(i, j);
format.backspace = false;
out.textContent = validate(this.value) ? "is valid" : "is invalid";
}
function validate(num) {
return /^BE [01](d{3}.){2}d{3}$/.test(num)
&& 97 - num.replace(/D/g, "").slice(0, 8) % 97 === +num.slice(-2);
}
ent.addEventListener("input", format);
ent.addEventListener("keydown", (e) => format.backspace = e.key == "Backspace");Belgian enterprise number: <input id="ent" value="BE ____.___.___">
<span id="isvalid"></span>Here is an implementation of a BE ____.___.___ style of input. The pattern will be maintained, so the input will be guaranteed to have the "BE" prefix, the space, and the two dots. The validation can then concentrate on completeness and the modulo test.
Note that the input requires the first group to have 4 digits, where the first digit must be a 0 or a 1.
const ent = document.getElementById("ent");
const out = document.getElementById("isvalid");
function format() {
const re = /^D*[2-9]+|D+/g;
const [i, j] = [this.selectionStart, this.selectionEnd].map(i => {
i = this.value.slice(0, i).replace(re, "").length;
return i + 3 + (i >= 4 + format.backspace) + (i >= 7 + format.backspace);
});
this.value = "BE " + this.value.replace(re, "").padEnd(10, "_")
.replace(/(....)(...)(...).*/, "$1.$2.$3");
this.setSelectionRange(i, j);
format.backspace = false;
out.textContent = validate(this.value) ? "is valid" : "is invalid";
}
function validate(num) {
return /^BE [01](d{3}.){2}d{3}$/.test(num)
&& 97 - num.replace(/D/g, "").slice(0, 8) % 97 === +num.slice(-2);
}
ent.addEventListener("input", format);
ent.addEventListener("keydown", (e) => format.backspace = e.key == "Backspace");Belgian enterprise number: <input id="ent" value="BE ____.___.___">
<span id="isvalid"></span>const ent = document.getElementById("ent");
const out = document.getElementById("isvalid");
function format() {
const re = /^D*[2-9]+|D+/g;
const [i, j] = [this.selectionStart, this.selectionEnd].map(i => {
i = this.value.slice(0, i).replace(re, "").length;
return i + 3 + (i >= 4 + format.backspace) + (i >= 7 + format.backspace);
});
this.value = "BE " + this.value.replace(re, "").padEnd(10, "_")
.replace(/(....)(...)(...).*/, "$1.$2.$3");
this.setSelectionRange(i, j);
format.backspace = false;
out.textContent = validate(this.value) ? "is valid" : "is invalid";
}
function validate(num) {
return /^BE [01](d{3}.){2}d{3}$/.test(num)
&& 97 - num.replace(/D/g, "").slice(0, 8) % 97 === +num.slice(-2);
}
ent.addEventListener("input", format);
ent.addEventListener("keydown", (e) => format.backspace = e.key == "Backspace");Belgian enterprise number: <input id="ent" value="BE ____.___.___">
<span id="isvalid"></span>const ent = document.getElementById("ent");
const out = document.getElementById("isvalid");
function format() {
const re = /^D*[2-9]+|D+/g;
const [i, j] = [this.selectionStart, this.selectionEnd].map(i => {
i = this.value.slice(0, i).replace(re, "").length;
return i + 3 + (i >= 4 + format.backspace) + (i >= 7 + format.backspace);
});
this.value = "BE " + this.value.replace(re, "").padEnd(10, "_")
.replace(/(....)(...)(...).*/, "$1.$2.$3");
this.setSelectionRange(i, j);
format.backspace = false;
out.textContent = validate(this.value) ? "is valid" : "is invalid";
}
function validate(num) {
return /^BE [01](d{3}.){2}d{3}$/.test(num)
&& 97 - num.replace(/D/g, "").slice(0, 8) % 97 === +num.slice(-2);
}
ent.addEventListener("input", format);
ent.addEventListener("keydown", (e) => format.backspace = e.key == "Backspace");Belgian enterprise number: <input id="ent" value="BE ____.___.___">
<span id="isvalid"></span>answered 41 mins ago
trincottrincot
126k1688122
126k1688122
Wow, that's a nice solution to the second question.
– Mouser
20 mins ago
add a comment |
Wow, that's a nice solution to the second question.
– Mouser
20 mins ago
Wow, that's a nice solution to the second question.
– Mouser
20 mins ago
Wow, that's a nice solution to the second question.
– Mouser
20 mins ago
add a comment |
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The modulo test is not easily possible in regex.
– trincot
3 hours ago