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Regex for certain words causes Spaces
How do I do a case-insensitive replace all for a string?Using RegEx Pattern Match to create DateTime fields in triggerRegEx in use with an InboundEmail.htmlBody String is not matching a PatternRegular Expression for words with spacesspecify column for PageBlockSection ItemPattern.Split doesn't behave as I've expectedHow to split tables in HTML string into string collection using apex?Issues with Regex matcher on APEXRead through small CSV file returns Regex too complicatedRegex implementation in Salesforce: getting from known string to end of the line
For instance i'm trying to search this string
"this is a search LLC which can end up having spaces"
using the below regex, it's working fine but when I match and then remove it's causes a bunch of white spaces, so I want to limit the number the white space to just 1, how could I go about adding that?
(?i)(?<=|^)LLC|end|having(?=|$)
this is a search LLC which can end up having spaces
this is a search which can up spaces
public static String removeValuesByRegex(String input, Pattern pattern, String replaceWith){
String result;
Matcher matcher = pattern.matcher(input);
return result = matcher.replaceAll(replaceWith).trim();
}
Then in my test class im doing
String keywords = 'LLC|end|Having';
Pattern pattern = Pattern.compile('(?i)(?<= |^)'|' + keyWords + ''|(?= |$)');
String stringResult = UtilString.removeValuesByRegex('This is a LLC and General Partnership Test ', pattern, '');
apex regular-expressions
add a comment |
For instance i'm trying to search this string
"this is a search LLC which can end up having spaces"
using the below regex, it's working fine but when I match and then remove it's causes a bunch of white spaces, so I want to limit the number the white space to just 1, how could I go about adding that?
(?i)(?<=|^)LLC|end|having(?=|$)
this is a search LLC which can end up having spaces
this is a search which can up spaces
public static String removeValuesByRegex(String input, Pattern pattern, String replaceWith){
String result;
Matcher matcher = pattern.matcher(input);
return result = matcher.replaceAll(replaceWith).trim();
}
Then in my test class im doing
String keywords = 'LLC|end|Having';
Pattern pattern = Pattern.compile('(?i)(?<= |^)'|' + keyWords + ''|(?= |$)');
String stringResult = UtilString.removeValuesByRegex('This is a LLC and General Partnership Test ', pattern, '');
apex regular-expressions
add a comment |
For instance i'm trying to search this string
"this is a search LLC which can end up having spaces"
using the below regex, it's working fine but when I match and then remove it's causes a bunch of white spaces, so I want to limit the number the white space to just 1, how could I go about adding that?
(?i)(?<=|^)LLC|end|having(?=|$)
this is a search LLC which can end up having spaces
this is a search which can up spaces
public static String removeValuesByRegex(String input, Pattern pattern, String replaceWith){
String result;
Matcher matcher = pattern.matcher(input);
return result = matcher.replaceAll(replaceWith).trim();
}
Then in my test class im doing
String keywords = 'LLC|end|Having';
Pattern pattern = Pattern.compile('(?i)(?<= |^)'|' + keyWords + ''|(?= |$)');
String stringResult = UtilString.removeValuesByRegex('This is a LLC and General Partnership Test ', pattern, '');
apex regular-expressions
For instance i'm trying to search this string
"this is a search LLC which can end up having spaces"
using the below regex, it's working fine but when I match and then remove it's causes a bunch of white spaces, so I want to limit the number the white space to just 1, how could I go about adding that?
(?i)(?<=|^)LLC|end|having(?=|$)
this is a search LLC which can end up having spaces
this is a search which can up spaces
public static String removeValuesByRegex(String input, Pattern pattern, String replaceWith){
String result;
Matcher matcher = pattern.matcher(input);
return result = matcher.replaceAll(replaceWith).trim();
}
Then in my test class im doing
String keywords = 'LLC|end|Having';
Pattern pattern = Pattern.compile('(?i)(?<= |^)'|' + keyWords + ''|(?= |$)');
String stringResult = UtilString.removeValuesByRegex('This is a LLC and General Partnership Test ', pattern, '');
apex regular-expressions
apex regular-expressions
asked 1 hour ago
EricSSHEricSSH
3,64211749
3,64211749
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
If you want to remove duplicate spacing, I recommend you call normalizeSpace
on the output:
String result = MyClass.myMethod(input, pattern, '').normalizeSpace();
Note from the documentation:
normalizeSpace()
Returns the current String with leading, trailing, and repeating white space characters removed.
Thanks, I missed that in the docs
– EricSSH
1 hour ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
If you want to remove duplicate spacing, I recommend you call normalizeSpace
on the output:
String result = MyClass.myMethod(input, pattern, '').normalizeSpace();
Note from the documentation:
normalizeSpace()
Returns the current String with leading, trailing, and repeating white space characters removed.
Thanks, I missed that in the docs
– EricSSH
1 hour ago
add a comment |
If you want to remove duplicate spacing, I recommend you call normalizeSpace
on the output:
String result = MyClass.myMethod(input, pattern, '').normalizeSpace();
Note from the documentation:
normalizeSpace()
Returns the current String with leading, trailing, and repeating white space characters removed.
Thanks, I missed that in the docs
– EricSSH
1 hour ago
add a comment |
If you want to remove duplicate spacing, I recommend you call normalizeSpace
on the output:
String result = MyClass.myMethod(input, pattern, '').normalizeSpace();
Note from the documentation:
normalizeSpace()
Returns the current String with leading, trailing, and repeating white space characters removed.
If you want to remove duplicate spacing, I recommend you call normalizeSpace
on the output:
String result = MyClass.myMethod(input, pattern, '').normalizeSpace();
Note from the documentation:
normalizeSpace()
Returns the current String with leading, trailing, and repeating white space characters removed.
answered 1 hour ago
Adrian Larson♦Adrian Larson
109k19115246
109k19115246
Thanks, I missed that in the docs
– EricSSH
1 hour ago
add a comment |
Thanks, I missed that in the docs
– EricSSH
1 hour ago
Thanks, I missed that in the docs
– EricSSH
1 hour ago
Thanks, I missed that in the docs
– EricSSH
1 hour ago
add a comment |
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