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How to merge and return new array from object in es6

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11

1

Suppose there are two objects.

const a = [

  { id: '1-1-1', name: 'a111' },

  { id: '1-1-2', name: 'a112' },

  { id: '1-2-1', name: 'a121' },

  { id: '1-2-2', name: 'a122' },

  { id: '2-1-1', name: 'a211' },

  { id: '2-1-2', name: 'a212' }

]

const b = ['1-1', '1-2', '2-1']

and the result

  {



      '1-1':[

        { id: '1-1-1', name: 'a111' },

        { id: '1-1-2', name: 'a112' },

      ],

      '1-2':[

          { id: '1-2-1', name: 'a121' },

          { id: '1-2-2', name: 'a122' },

      ],

      '2-1':[

        { id: '2-1-1', name: 'a211' },

        { id: '2-1-2', name: 'a212' },

      ]

    }

Basically, I want to group the data.

I use includes to check if the item from b to match the id from a. Then construct the new array.

This is my attempt(fiddle):

return b.map(item => a.map(jtem => {

  if(jtem.id.includes(item)){

    return {

      [item]: jtem

    }

  }

}))

For somehow, it doesn't work.

and, is there a clever way to avoid the nested for loop or map function?

javascript ecmascript-6 ecmascript-7

share|improve this question

edited 23 hours ago

SPG

asked 23 hours ago

SPGSPG

2,439103359

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Shouldn't the result be an object?
  
  – Jack Bashford
  23 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @JackBashford Hey man, sry, u r right, I just updated it.
  
  – SPG
  23 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Do you really want includes? I'd recommend startsWith
  
  – Bergi
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Bergi thx, I think startWith is better
  
  – SPG
  5 hours ago
  

  
  
  
  

add a comment | 

11

1

Suppose there are two objects.

const a = [

  { id: '1-1-1', name: 'a111' },

  { id: '1-1-2', name: 'a112' },

  { id: '1-2-1', name: 'a121' },

  { id: '1-2-2', name: 'a122' },

  { id: '2-1-1', name: 'a211' },

  { id: '2-1-2', name: 'a212' }

]

const b = ['1-1', '1-2', '2-1']

and the result

  {



      '1-1':[

        { id: '1-1-1', name: 'a111' },

        { id: '1-1-2', name: 'a112' },

      ],

      '1-2':[

          { id: '1-2-1', name: 'a121' },

          { id: '1-2-2', name: 'a122' },

      ],

      '2-1':[

        { id: '2-1-1', name: 'a211' },

        { id: '2-1-2', name: 'a212' },

      ]

    }

Basically, I want to group the data.

I use includes to check if the item from b to match the id from a. Then construct the new array.

This is my attempt(fiddle):

return b.map(item => a.map(jtem => {

  if(jtem.id.includes(item)){

    return {

      [item]: jtem

    }

  }

}))

For somehow, it doesn't work.

and, is there a clever way to avoid the nested for loop or map function?

javascript ecmascript-6 ecmascript-7

share|improve this question

edited 23 hours ago

SPG

asked 23 hours ago

SPGSPG

2,439103359

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Shouldn't the result be an object?
  
  – Jack Bashford
  23 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @JackBashford Hey man, sry, u r right, I just updated it.
  
  – SPG
  23 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Do you really want includes? I'd recommend startsWith
  
  – Bergi
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Bergi thx, I think startWith is better
  
  – SPG
  5 hours ago
  

  
  
  
  

add a comment | 

Suppose there are two objects.

const a = [

  { id: '1-1-1', name: 'a111' },

  { id: '1-1-2', name: 'a112' },

  { id: '1-2-1', name: 'a121' },

  { id: '1-2-2', name: 'a122' },

  { id: '2-1-1', name: 'a211' },

  { id: '2-1-2', name: 'a212' }

]

const b = ['1-1', '1-2', '2-1']

and the result

  {



      '1-1':[

        { id: '1-1-1', name: 'a111' },

        { id: '1-1-2', name: 'a112' },

      ],

      '1-2':[

          { id: '1-2-1', name: 'a121' },

          { id: '1-2-2', name: 'a122' },

      ],

      '2-1':[

        { id: '2-1-1', name: 'a211' },

        { id: '2-1-2', name: 'a212' },

      ]

    }

Basically, I want to group the data.

I use includes to check if the item from b to match the id from a. Then construct the new array.

This is my attempt(fiddle):

return b.map(item => a.map(jtem => {

  if(jtem.id.includes(item)){

    return {

      [item]: jtem

    }

  }

}))

For somehow, it doesn't work.

and, is there a clever way to avoid the nested for loop or map function?

javascript ecmascript-6 ecmascript-7

share|improve this question

edited 23 hours ago

SPG

asked 23 hours ago

SPGSPG

2,439103359

const a = [

  { id: '1-1-1', name: 'a111' },

  { id: '1-1-2', name: 'a112' },

  { id: '1-2-1', name: 'a121' },

  { id: '1-2-2', name: 'a122' },

  { id: '2-1-1', name: 'a211' },

  { id: '2-1-2', name: 'a212' }

]

const b = ['1-1', '1-2', '2-1']

  {



      '1-1':[

        { id: '1-1-1', name: 'a111' },

        { id: '1-1-2', name: 'a112' },

      ],

      '1-2':[

          { id: '1-2-1', name: 'a121' },

          { id: '1-2-2', name: 'a122' },

      ],

      '2-1':[

        { id: '2-1-1', name: 'a211' },

        { id: '2-1-2', name: 'a212' },

      ]

    }

return b.map(item => a.map(jtem => {

  if(jtem.id.includes(item)){

    return {

      [item]: jtem

    }

  }

}))

share|improve this question

edited 23 hours ago

SPG

asked 23 hours ago

SPGSPG

2,439103359

share|improve this question

edited 23 hours ago

SPG

asked 23 hours ago

SPGSPG

2,439103359

asked 23 hours ago

SPGSPG

2,439103359

asked 23 hours ago

SPGSPG

2,439103359

1 Answer
1

active

oldest

votes

14

You can do that in following steps:

• Apply reduce() on the array b




• During each iteration use filter() on the the array a



• Get all the items from a which starts with item of b using String.prototype.startsWith()
  


• At last set it as property of the ac and return ac
  

const a = [

  { id: '1-1-1', name: 'a111' },

  { id: '1-1-2', name: 'a112' },

  { id: '1-2-1', name: 'a121' },

  { id: '1-2-2', name: 'a122' },

  { id: '2-1-1', name: 'a211' },

  { id: '2-1-2', name: 'a212' }

]

const b = ['1-1', '1-2', '2-1']



let res = b.reduce((ac,b) => {

  

  ac[b] = a.filter(x => x.id.startsWith(b));

  return ac;



},{})

console.log(res)

As suggested by @Falco is the comments that It would be better to scan over the a once as its large. So here is that version.Actually its better regarding performance

const a = [

  { id: '1-1-1', name: 'a111' },

  { id: '1-1-2', name: 'a112' },

  { id: '1-2-1', name: 'a121' },

  { id: '1-2-2', name: 'a122' },

  { id: '2-1-1', name: 'a211' },

  { id: '2-1-2', name: 'a212' }

]

const b = ['1-1', '1-2', '2-1']





let res = a.reduce((ac,x) => {

  let temp = b.find(y => x.id.startsWith(y))

  if(!ac[temp]) ac[temp] = [];

  ac[temp].push(x);

  return ac;

},{})



console.log(res)

Note: startsWith is not supported by I.E. So you can create polyfill using indexOf

if(!String.prototype.startWith){

  String.prototype.startsWith = function(str){

    return this.indexOf(str) === 0

  }

}

share|improve this answer

edited 14 hours ago

answered 23 hours ago

Maheer AliMaheer Ali

7,451619

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  While it is specifically said in the question that the op wants to use es6, and that IE don't support es6 features, I just want to mention that startsWith() don't work in IE (while reduce, filter, and setting a property of an object is totally fine if IE > 9) and if someone wants to do the same thing that startsWith do, they can implment their own with some substring :)
  
  – Neyt
  20 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  For big a and small b I would probably go with a.reduce(...) because of locality and only scan over the big array once.
  
  – Falco
  18 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Falco Thanks for suggestion I updated.
  
  – Maheer Ali
  18 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @Neyt Thanks for suggestion I updated
  
  – Maheer Ali
  18 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @MaheerAli Thank you - here is a benchmark comparing the two :-) jsbench.me/dfjtoadysr/1
  
  – Falco
  18 hours ago
  

  
  
  
  

add a comment | 

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14

You can do that in following steps:

• Apply reduce() on the array b




• During each iteration use filter() on the the array a



• Get all the items from a which starts with item of b using String.prototype.startsWith()
  


• At last set it as property of the ac and return ac
  

const a = [

  { id: '1-1-1', name: 'a111' },

  { id: '1-1-2', name: 'a112' },

  { id: '1-2-1', name: 'a121' },

  { id: '1-2-2', name: 'a122' },

  { id: '2-1-1', name: 'a211' },

  { id: '2-1-2', name: 'a212' }

]

const b = ['1-1', '1-2', '2-1']



let res = b.reduce((ac,b) => {

  

  ac[b] = a.filter(x => x.id.startsWith(b));

  return ac;



},{})

console.log(res)

As suggested by @Falco is the comments that It would be better to scan over the a once as its large. So here is that version.Actually its better regarding performance

const a = [

  { id: '1-1-1', name: 'a111' },

  { id: '1-1-2', name: 'a112' },

  { id: '1-2-1', name: 'a121' },

  { id: '1-2-2', name: 'a122' },

  { id: '2-1-1', name: 'a211' },

  { id: '2-1-2', name: 'a212' }

]

const b = ['1-1', '1-2', '2-1']





let res = a.reduce((ac,x) => {

  let temp = b.find(y => x.id.startsWith(y))

  if(!ac[temp]) ac[temp] = [];

  ac[temp].push(x);

  return ac;

},{})



console.log(res)

Note: startsWith is not supported by I.E. So you can create polyfill using indexOf

if(!String.prototype.startWith){

  String.prototype.startsWith = function(str){

    return this.indexOf(str) === 0

  }

}

share|improve this answer

edited 14 hours ago

answered 23 hours ago

Maheer AliMaheer Ali

7,451619

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  While it is specifically said in the question that the op wants to use es6, and that IE don't support es6 features, I just want to mention that startsWith() don't work in IE (while reduce, filter, and setting a property of an object is totally fine if IE > 9) and if someone wants to do the same thing that startsWith do, they can implment their own with some substring :)
  
  – Neyt
  20 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  For big a and small b I would probably go with a.reduce(...) because of locality and only scan over the big array once.
  
  – Falco
  18 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Falco Thanks for suggestion I updated.
  
  – Maheer Ali
  18 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @Neyt Thanks for suggestion I updated
  
  – Maheer Ali
  18 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @MaheerAli Thank you - here is a benchmark comparing the two :-) jsbench.me/dfjtoadysr/1
  
  – Falco
  18 hours ago
  

  
  
  
  

add a comment | 

14

You can do that in following steps:

• Apply reduce() on the array b




• During each iteration use filter() on the the array a



• Get all the items from a which starts with item of b using String.prototype.startsWith()
  


• At last set it as property of the ac and return ac
  

const a = [

  { id: '1-1-1', name: 'a111' },

  { id: '1-1-2', name: 'a112' },

  { id: '1-2-1', name: 'a121' },

  { id: '1-2-2', name: 'a122' },

  { id: '2-1-1', name: 'a211' },

  { id: '2-1-2', name: 'a212' }

]

const b = ['1-1', '1-2', '2-1']



let res = b.reduce((ac,b) => {

  

  ac[b] = a.filter(x => x.id.startsWith(b));

  return ac;



},{})

console.log(res)

As suggested by @Falco is the comments that It would be better to scan over the a once as its large. So here is that version.Actually its better regarding performance

const a = [

  { id: '1-1-1', name: 'a111' },

  { id: '1-1-2', name: 'a112' },

  { id: '1-2-1', name: 'a121' },

  { id: '1-2-2', name: 'a122' },

  { id: '2-1-1', name: 'a211' },

  { id: '2-1-2', name: 'a212' }

]

const b = ['1-1', '1-2', '2-1']





let res = a.reduce((ac,x) => {

  let temp = b.find(y => x.id.startsWith(y))

  if(!ac[temp]) ac[temp] = [];

  ac[temp].push(x);

  return ac;

},{})



console.log(res)

Note: startsWith is not supported by I.E. So you can create polyfill using indexOf

if(!String.prototype.startWith){

  String.prototype.startsWith = function(str){

    return this.indexOf(str) === 0

  }

}

share|improve this answer

edited 14 hours ago

answered 23 hours ago

Maheer AliMaheer Ali

7,451619

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  While it is specifically said in the question that the op wants to use es6, and that IE don't support es6 features, I just want to mention that startsWith() don't work in IE (while reduce, filter, and setting a property of an object is totally fine if IE > 9) and if someone wants to do the same thing that startsWith do, they can implment their own with some substring :)
  
  – Neyt
  20 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  For big a and small b I would probably go with a.reduce(...) because of locality and only scan over the big array once.
  
  – Falco
  18 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Falco Thanks for suggestion I updated.
  
  – Maheer Ali
  18 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @Neyt Thanks for suggestion I updated
  
  – Maheer Ali
  18 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @MaheerAli Thank you - here is a benchmark comparing the two :-) jsbench.me/dfjtoadysr/1
  
  – Falco
  18 hours ago
  

  
  
  
  

add a comment | 

You can do that in following steps:

• Apply reduce() on the array b




• During each iteration use filter() on the the array a



• Get all the items from a which starts with item of b using String.prototype.startsWith()
  


• At last set it as property of the ac and return ac
  

const a = [

  { id: '1-1-1', name: 'a111' },

  { id: '1-1-2', name: 'a112' },

  { id: '1-2-1', name: 'a121' },

  { id: '1-2-2', name: 'a122' },

  { id: '2-1-1', name: 'a211' },

  { id: '2-1-2', name: 'a212' }

]

const b = ['1-1', '1-2', '2-1']



let res = b.reduce((ac,b) => {

  

  ac[b] = a.filter(x => x.id.startsWith(b));

  return ac;



},{})

console.log(res)

As suggested by @Falco is the comments that It would be better to scan over the a once as its large. So here is that version.Actually its better regarding performance

const a = [

  { id: '1-1-1', name: 'a111' },

  { id: '1-1-2', name: 'a112' },

  { id: '1-2-1', name: 'a121' },

  { id: '1-2-2', name: 'a122' },

  { id: '2-1-1', name: 'a211' },

  { id: '2-1-2', name: 'a212' }

]

const b = ['1-1', '1-2', '2-1']





let res = a.reduce((ac,x) => {

  let temp = b.find(y => x.id.startsWith(y))

  if(!ac[temp]) ac[temp] = [];

  ac[temp].push(x);

  return ac;

},{})



console.log(res)

Note: startsWith is not supported by I.E. So you can create polyfill using indexOf

if(!String.prototype.startWith){

  String.prototype.startsWith = function(str){

    return this.indexOf(str) === 0

  }

}

share|improve this answer

edited 14 hours ago

answered 23 hours ago

Maheer AliMaheer Ali

7,451619

const a = [

  { id: '1-1-1', name: 'a111' },

  { id: '1-1-2', name: 'a112' },

  { id: '1-2-1', name: 'a121' },

  { id: '1-2-2', name: 'a122' },

  { id: '2-1-1', name: 'a211' },

  { id: '2-1-2', name: 'a212' }

]

const b = ['1-1', '1-2', '2-1']



let res = b.reduce((ac,b) => {

  

  ac[b] = a.filter(x => x.id.startsWith(b));

  return ac;



},{})

console.log(res)

const a = [

  { id: '1-1-1', name: 'a111' },

  { id: '1-1-2', name: 'a112' },

  { id: '1-2-1', name: 'a121' },

  { id: '1-2-2', name: 'a122' },

  { id: '2-1-1', name: 'a211' },

  { id: '2-1-2', name: 'a212' }

]

const b = ['1-1', '1-2', '2-1']





let res = a.reduce((ac,x) => {

  let temp = b.find(y => x.id.startsWith(y))

  if(!ac[temp]) ac[temp] = [];

  ac[temp].push(x);

  return ac;

},{})



console.log(res)

if(!String.prototype.startWith){

  String.prototype.startsWith = function(str){

    return this.indexOf(str) === 0

  }

}

const a = [

  { id: '1-1-1', name: 'a111' },

  { id: '1-1-2', name: 'a112' },

  { id: '1-2-1', name: 'a121' },

  { id: '1-2-2', name: 'a122' },

  { id: '2-1-1', name: 'a211' },

  { id: '2-1-2', name: 'a212' }

]

const b = ['1-1', '1-2', '2-1']



let res = b.reduce((ac,b) => {

  

  ac[b] = a.filter(x => x.id.startsWith(b));

  return ac;



},{})

console.log(res)

const a = [

  { id: '1-1-1', name: 'a111' },

  { id: '1-1-2', name: 'a112' },

  { id: '1-2-1', name: 'a121' },

  { id: '1-2-2', name: 'a122' },

  { id: '2-1-1', name: 'a211' },

  { id: '2-1-2', name: 'a212' }

]

const b = ['1-1', '1-2', '2-1']



let res = b.reduce((ac,b) => {

  

  ac[b] = a.filter(x => x.id.startsWith(b));

  return ac;



},{})

console.log(res)

const a = [

  { id: '1-1-1', name: 'a111' },

  { id: '1-1-2', name: 'a112' },

  { id: '1-2-1', name: 'a121' },

  { id: '1-2-2', name: 'a122' },

  { id: '2-1-1', name: 'a211' },

  { id: '2-1-2', name: 'a212' }

]

const b = ['1-1', '1-2', '2-1']





let res = a.reduce((ac,x) => {

  let temp = b.find(y => x.id.startsWith(y))

  if(!ac[temp]) ac[temp] = [];

  ac[temp].push(x);

  return ac;

},{})



console.log(res)

const a = [

  { id: '1-1-1', name: 'a111' },

  { id: '1-1-2', name: 'a112' },

  { id: '1-2-1', name: 'a121' },

  { id: '1-2-2', name: 'a122' },

  { id: '2-1-1', name: 'a211' },

  { id: '2-1-2', name: 'a212' }

]

const b = ['1-1', '1-2', '2-1']





let res = a.reduce((ac,x) => {

  let temp = b.find(y => x.id.startsWith(y))

  if(!ac[temp]) ac[temp] = [];

  ac[temp].push(x);

  return ac;

},{})



console.log(res)

if(!String.prototype.startWith){

  String.prototype.startsWith = function(str){

    return this.indexOf(str) === 0

  }

}

if(!String.prototype.startWith){

  String.prototype.startsWith = function(str){

    return this.indexOf(str) === 0

  }

}

share|improve this answer

edited 14 hours ago

answered 23 hours ago

Maheer AliMaheer Ali

7,451619

answered 23 hours ago

Maheer AliMaheer Ali

7,451619

answered 23 hours ago

Maheer AliMaheer Ali

7,451619