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JavaScript array of objects contains the same array data

7

1

I try to get all same data values into an array of objects. This is my input:

var a = [
 name: "Foo",
 id: "123",
 data: ["65d4ze", "65h8914d"]
 ,
 
 name: "Bar",
 id: "321",
 data: ["65d4ze", "894ver81"]
 
]

I need a result like:

["65d4ze"]

I try to loop on my object to get this output, but I'm completely lost... I don't know how to know if the result is into all data arrays.

var a = [
 name: "Foo",
 id: "123",
 data: ["65d4ze", "65h8914d"]
 ,
 
 name: "Bar",
 id: "321",
 data: ["65d4ze", "894ver81"]
 
 ],
 b = [],
 c = []
a.forEach(function(object) 
 b.push(object.data.map(function(val) 
 return val;
 )
 );
);

console.log(b);

javascript arrays

share|improve this question

edited Mar 25 at 12:05

Peter Mortensen

13.8k1987113

asked Mar 25 at 10:04

KamouloxKamoulox

533

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  So get all arrays elements that are present on all a?
  
  – Eddie
  Mar 25 at 10:06
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  Possible duplicate of Simplest code for array intersection in javascript
  
  – James Long
  Mar 25 at 10:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @JamesLong I don't understand the duplication, why this answer can help me ?
  
  – Kamoulox
  Mar 25 at 10:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Eddie I need to get all array into the key data. This key is present into all object of my principal array.
  
  – Kamoulox
  Mar 25 at 10:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Kamoulox you have 2 arrays (a[0].data and a[1].data). You seem to be wanting an array of all the values that appear in both arrays. Unless I'm mis-reading your question
  
  – James Long
  Mar 25 at 10:12
  

  
  
  
  

 | 
show 3 more comments

7

1

I try to get all same data values into an array of objects. This is my input:

var a = [
 name: "Foo",
 id: "123",
 data: ["65d4ze", "65h8914d"]
 ,
 
 name: "Bar",
 id: "321",
 data: ["65d4ze", "894ver81"]
 
]

I need a result like:

["65d4ze"]

I try to loop on my object to get this output, but I'm completely lost... I don't know how to know if the result is into all data arrays.

var a = [
 name: "Foo",
 id: "123",
 data: ["65d4ze", "65h8914d"]
 ,
 
 name: "Bar",
 id: "321",
 data: ["65d4ze", "894ver81"]
 
 ],
 b = [],
 c = []
a.forEach(function(object) 
 b.push(object.data.map(function(val) 
 return val;
 )
 );
);

console.log(b);

javascript arrays

share|improve this question

edited Mar 25 at 12:05

Peter Mortensen

13.8k1987113

asked Mar 25 at 10:04

KamouloxKamoulox

533

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  So get all arrays elements that are present on all a?
  
  – Eddie
  Mar 25 at 10:06
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  Possible duplicate of Simplest code for array intersection in javascript
  
  – James Long
  Mar 25 at 10:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @JamesLong I don't understand the duplication, why this answer can help me ?
  
  – Kamoulox
  Mar 25 at 10:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Eddie I need to get all array into the key data. This key is present into all object of my principal array.
  
  – Kamoulox
  Mar 25 at 10:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Kamoulox you have 2 arrays (a[0].data and a[1].data). You seem to be wanting an array of all the values that appear in both arrays. Unless I'm mis-reading your question
  
  – James Long
  Mar 25 at 10:12
  

  
  
  
  

 | 
show 3 more comments

I try to get all same data values into an array of objects. This is my input:

var a = [
 name: "Foo",
 id: "123",
 data: ["65d4ze", "65h8914d"]
 ,
 
 name: "Bar",
 id: "321",
 data: ["65d4ze", "894ver81"]
 
]

I need a result like:

["65d4ze"]

I try to loop on my object to get this output, but I'm completely lost... I don't know how to know if the result is into all data arrays.

var a = [
 name: "Foo",
 id: "123",
 data: ["65d4ze", "65h8914d"]
 ,
 
 name: "Bar",
 id: "321",
 data: ["65d4ze", "894ver81"]
 
 ],
 b = [],
 c = []
a.forEach(function(object) 
 b.push(object.data.map(function(val) 
 return val;
 )
 );
);

console.log(b);

javascript arrays

share|improve this question

edited Mar 25 at 12:05

Peter Mortensen

13.8k1987113

asked Mar 25 at 10:04

KamouloxKamoulox

533

var a = [
 name: "Foo",
 id: "123",
 data: ["65d4ze", "65h8914d"]
 ,
 
 name: "Bar",
 id: "321",
 data: ["65d4ze", "894ver81"]
 
]

["65d4ze"]

var a = [
 name: "Foo",
 id: "123",
 data: ["65d4ze", "65h8914d"]
 ,
 
 name: "Bar",
 id: "321",
 data: ["65d4ze", "894ver81"]
 
 ],
 b = [],
 c = []
a.forEach(function(object) 
 b.push(object.data.map(function(val) 
 return val;
 )
 );
);

console.log(b);

var a = [
 name: "Foo",
 id: "123",
 data: ["65d4ze", "65h8914d"]
 ,
 
 name: "Bar",
 id: "321",
 data: ["65d4ze", "894ver81"]
 
 ],
 b = [],
 c = []
a.forEach(function(object) 
 b.push(object.data.map(function(val) 
 return val;
 )
 );
);

console.log(b);

var a = [
 name: "Foo",
 id: "123",
 data: ["65d4ze", "65h8914d"]
 ,
 
 name: "Bar",
 id: "321",
 data: ["65d4ze", "894ver81"]
 
 ],
 b = [],
 c = []
a.forEach(function(object) 
 b.push(object.data.map(function(val) 
 return val;
 )
 );
);

console.log(b);

share|improve this question

edited Mar 25 at 12:05

Peter Mortensen

13.8k1987113

asked Mar 25 at 10:04

KamouloxKamoulox

533

share|improve this question

edited Mar 25 at 12:05

Peter Mortensen

13.8k1987113

asked Mar 25 at 10:04

KamouloxKamoulox

533

edited Mar 25 at 12:05

Peter Mortensen

13.8k1987113

edited Mar 25 at 12:05

Peter Mortensen

13.8k1987113

asked Mar 25 at 10:04

KamouloxKamoulox

533

asked Mar 25 at 10:04

KamouloxKamoulox

533

5 Answers
5

active

oldest

votes

12

You could map data and get the common values with Array#map, Array#reduce, Array#filter, Set and Set#has.

var array = [ name: "Foo", id: "123", data: ["65d4ze", "65h8914d"] , name: "Bar", id: "321", data: ["65d4ze", "894ver81"] ],
 key = 'data',
 common = array
 .map(o => o[key])
 .reduce((a, b) => b.filter(Set.prototype.has, new Set(a)));

console.log(common);

share|improve this answer

answered Mar 25 at 10:09

Nina ScholzNina Scholz

195k15107179

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I have some troubles to understand this piece of code b.filter(Set.prototype.has, new Set(a)). But it's working perfectly, thanks
  
  – Kamoulox
  Mar 25 at 10:16
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  filter has two parameters, one foir the callback, which is here a prototype function of Set, has which checks a value against a set. at this point, the method does not works because it need an instance of Set. the second parameter is thisArg, where you can hand over an object, which is used as this in the callback. mabe a different use makes ir a bet more clear. you could gate the same with a bound this by using this as callback: Set.prototype.has.bind(new Set(a)), which is now a function with this.
  
  – Nina Scholz
  Mar 25 at 10:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  With array common = array .map(o => o[key]) .reduce((a, b) => b.filter(Array.prototype.includes, a));
  
  – Pranav C Balan
  Mar 25 at 10:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @PranavCBalan, Array#includes uses a second parameter fromIndex and gets this value with the index of the callback ...
  
  – Nina Scholz
  Mar 25 at 10:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @NinaScholz : oops that's right, that would make issues :)
  
  – Pranav C Balan
  Mar 25 at 10:25
  
  

  
  
  
  

add a comment | 

3

You can use the Array#filter method. Filter the first array by checking if a value is present in all other object properties (arrays), using the Array#every method to check if a value is present in all remaining arrays.

let res = a[0].data.filter(v => a.slice(1).every(a => a.data.includes(v)));

var a = [
 name: "Foo",
 id: "123",
 data: ["65d4ze", "65h8914d"]
 ,
 
 name: "Bar",
 id: "321",
 data: ["65d4ze", "894ver81"]
 
];

let res = a[0].data.filter(v => a.slice(1).every(a => a.data.includes(v)));

console.log(res)

share|improve this answer

edited Mar 25 at 12:28

Peter Mortensen

13.8k1987113

answered Mar 25 at 10:14

Pranav C BalanPranav C Balan

88.6k1391117

add a comment | 

1

var a = [
 name: "Foo",
 id: "123",
 data: ["65d4ze", "65h8914d"]
 ,
 
 name: "Bar",
 id: "321",
 data: ["65d4ze", "894ver81"]
 
 ],
 b = ;
a.forEach(function(i) 
 i.data.forEach(function(j) 
 if (!b.hasOwnProperty(j)) 
 b[j] = 0;
 
 b[j] = b[j] + 1;
 );
);
c = []
for (var i in b) 
 if (b.hasOwnProperty(i)) 
 if (b[i] > 1) 
 c.push(i)
 
 

console.log(c);

share|improve this answer

answered Mar 25 at 10:12

Shree TiwariShree Tiwari

287112

add a comment | 

1

Use the flat function in the array:

var a = [
 name: "Foo",
 id: "123",
 data: ["65d4ze", "65h8914d"]
 ,
 
 name: "Bar",
 id: "321",
 data: ["65d4ze", "894ver81"]
 
 ],
 b = [],
 c = []
a.forEach(function(object) 
 b.push(object.data.map(function(val) 
 return val;
 )
 );
);

console.log(b.flat());

share|improve this answer

edited Mar 25 at 12:29

Peter Mortensen

13.8k1987113

answered Mar 25 at 10:18

thelastwormthelastworm

1428

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks for your answer, that a great code. But flat is not working on my browser
  
  – Kamoulox
  Mar 25 at 10:21
  

  
  
  
  

add a comment | 

1

You could use reduce and concat on each data array, and check the count of each item.

In the end, you check whether all objects across the array contain that item and return it if yes.

Note that this function works if you want to extract the item that has the same occurrence across all objects in the array.

If an item has duplicates, but does not fulfill the above condition, it would not be extracted.

let a = [name: "Foo",id: "123",data: ["65d4ze", "65h8914d"],name: "Bar",id: "321",data: ["65d4ze", "894ver81"]]

let arr = a.reduce((prev,next) => prev.data.concat(next.data))
let counts = ;
let result = [];
for (var i = 0; i < arr.length; i++) 
 var num = arr[i];
 counts[num] = counts[num] ? counts[num] + 1 : 1;


for (let i in counts) 
 if (counts[i] === a.length) 
 result.push(i)
 

console.log(result)

share|improve this answer

edited Mar 25 at 12:30

Peter Mortensen

13.8k1987113

answered Mar 25 at 10:55

tnkhtnkh

204110

add a comment | 

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12

You could map data and get the common values with Array#map, Array#reduce, Array#filter, Set and Set#has.

var array = [ name: "Foo", id: "123", data: ["65d4ze", "65h8914d"] , name: "Bar", id: "321", data: ["65d4ze", "894ver81"] ],
 key = 'data',
 common = array
 .map(o => o[key])
 .reduce((a, b) => b.filter(Set.prototype.has, new Set(a)));

console.log(common);

share|improve this answer

answered Mar 25 at 10:09

Nina ScholzNina Scholz

195k15107179

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I have some troubles to understand this piece of code b.filter(Set.prototype.has, new Set(a)). But it's working perfectly, thanks
  
  – Kamoulox
  Mar 25 at 10:16
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  filter has two parameters, one foir the callback, which is here a prototype function of Set, has which checks a value against a set. at this point, the method does not works because it need an instance of Set. the second parameter is thisArg, where you can hand over an object, which is used as this in the callback. mabe a different use makes ir a bet more clear. you could gate the same with a bound this by using this as callback: Set.prototype.has.bind(new Set(a)), which is now a function with this.
  
  – Nina Scholz
  Mar 25 at 10:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  With array common = array .map(o => o[key]) .reduce((a, b) => b.filter(Array.prototype.includes, a));
  
  – Pranav C Balan
  Mar 25 at 10:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @PranavCBalan, Array#includes uses a second parameter fromIndex and gets this value with the index of the callback ...
  
  – Nina Scholz
  Mar 25 at 10:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @NinaScholz : oops that's right, that would make issues :)
  
  – Pranav C Balan
  Mar 25 at 10:25
  
  

  
  
  
  

add a comment | 

12

You could map data and get the common values with Array#map, Array#reduce, Array#filter, Set and Set#has.

var array = [ name: "Foo", id: "123", data: ["65d4ze", "65h8914d"] , name: "Bar", id: "321", data: ["65d4ze", "894ver81"] ],
 key = 'data',
 common = array
 .map(o => o[key])
 .reduce((a, b) => b.filter(Set.prototype.has, new Set(a)));

console.log(common);

share|improve this answer

answered Mar 25 at 10:09

Nina ScholzNina Scholz

195k15107179

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I have some troubles to understand this piece of code b.filter(Set.prototype.has, new Set(a)). But it's working perfectly, thanks
  
  – Kamoulox
  Mar 25 at 10:16
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  filter has two parameters, one foir the callback, which is here a prototype function of Set, has which checks a value against a set. at this point, the method does not works because it need an instance of Set. the second parameter is thisArg, where you can hand over an object, which is used as this in the callback. mabe a different use makes ir a bet more clear. you could gate the same with a bound this by using this as callback: Set.prototype.has.bind(new Set(a)), which is now a function with this.
  
  – Nina Scholz
  Mar 25 at 10:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  With array common = array .map(o => o[key]) .reduce((a, b) => b.filter(Array.prototype.includes, a));
  
  – Pranav C Balan
  Mar 25 at 10:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @PranavCBalan, Array#includes uses a second parameter fromIndex and gets this value with the index of the callback ...
  
  – Nina Scholz
  Mar 25 at 10:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @NinaScholz : oops that's right, that would make issues :)
  
  – Pranav C Balan
  Mar 25 at 10:25
  
  

  
  
  
  

add a comment | 

You could map data and get the common values with Array#map, Array#reduce, Array#filter, Set and Set#has.

var array = [ name: "Foo", id: "123", data: ["65d4ze", "65h8914d"] , name: "Bar", id: "321", data: ["65d4ze", "894ver81"] ],
 key = 'data',
 common = array
 .map(o => o[key])
 .reduce((a, b) => b.filter(Set.prototype.has, new Set(a)));

console.log(common);

share|improve this answer

answered Mar 25 at 10:09

Nina ScholzNina Scholz

195k15107179

var array = [ name: "Foo", id: "123", data: ["65d4ze", "65h8914d"] , name: "Bar", id: "321", data: ["65d4ze", "894ver81"] ],
 key = 'data',
 common = array
 .map(o => o[key])
 .reduce((a, b) => b.filter(Set.prototype.has, new Set(a)));

console.log(common);

var array = [ name: "Foo", id: "123", data: ["65d4ze", "65h8914d"] , name: "Bar", id: "321", data: ["65d4ze", "894ver81"] ],
 key = 'data',
 common = array
 .map(o => o[key])
 .reduce((a, b) => b.filter(Set.prototype.has, new Set(a)));

console.log(common);

var array = [ name: "Foo", id: "123", data: ["65d4ze", "65h8914d"] , name: "Bar", id: "321", data: ["65d4ze", "894ver81"] ],
 key = 'data',
 common = array
 .map(o => o[key])
 .reduce((a, b) => b.filter(Set.prototype.has, new Set(a)));

console.log(common);

share|improve this answer

answered Mar 25 at 10:09

Nina ScholzNina Scholz

195k15107179

answered Mar 25 at 10:09

Nina ScholzNina Scholz

195k15107179

answered Mar 25 at 10:09

Nina ScholzNina Scholz

195k15107179

3

You can use the Array#filter method. Filter the first array by checking if a value is present in all other object properties (arrays), using the Array#every method to check if a value is present in all remaining arrays.

let res = a[0].data.filter(v => a.slice(1).every(a => a.data.includes(v)));

var a = [
 name: "Foo",
 id: "123",
 data: ["65d4ze", "65h8914d"]
 ,
 
 name: "Bar",
 id: "321",
 data: ["65d4ze", "894ver81"]
 
];

let res = a[0].data.filter(v => a.slice(1).every(a => a.data.includes(v)));

console.log(res)

share|improve this answer

edited Mar 25 at 12:28

Peter Mortensen

13.8k1987113

answered Mar 25 at 10:14

Pranav C BalanPranav C Balan

88.6k1391117

add a comment | 

3

You can use the Array#filter method. Filter the first array by checking if a value is present in all other object properties (arrays), using the Array#every method to check if a value is present in all remaining arrays.

let res = a[0].data.filter(v => a.slice(1).every(a => a.data.includes(v)));

var a = [
 name: "Foo",
 id: "123",
 data: ["65d4ze", "65h8914d"]
 ,
 
 name: "Bar",
 id: "321",
 data: ["65d4ze", "894ver81"]
 
];

let res = a[0].data.filter(v => a.slice(1).every(a => a.data.includes(v)));

console.log(res)

share|improve this answer

edited Mar 25 at 12:28

Peter Mortensen

13.8k1987113

answered Mar 25 at 10:14

Pranav C BalanPranav C Balan

88.6k1391117

add a comment | 

You can use the Array#filter method. Filter the first array by checking if a value is present in all other object properties (arrays), using the Array#every method to check if a value is present in all remaining arrays.

let res = a[0].data.filter(v => a.slice(1).every(a => a.data.includes(v)));

var a = [
 name: "Foo",
 id: "123",
 data: ["65d4ze", "65h8914d"]
 ,
 
 name: "Bar",
 id: "321",
 data: ["65d4ze", "894ver81"]
 
];

let res = a[0].data.filter(v => a.slice(1).every(a => a.data.includes(v)));

console.log(res)

share|improve this answer

edited Mar 25 at 12:28

Peter Mortensen

13.8k1987113

answered Mar 25 at 10:14

Pranav C BalanPranav C Balan

88.6k1391117

let res = a[0].data.filter(v => a.slice(1).every(a => a.data.includes(v)));

var a = [
 name: "Foo",
 id: "123",
 data: ["65d4ze", "65h8914d"]
 ,
 
 name: "Bar",
 id: "321",
 data: ["65d4ze", "894ver81"]
 
];

let res = a[0].data.filter(v => a.slice(1).every(a => a.data.includes(v)));

console.log(res)

var a = [
 name: "Foo",
 id: "123",
 data: ["65d4ze", "65h8914d"]
 ,
 
 name: "Bar",
 id: "321",
 data: ["65d4ze", "894ver81"]
 
];

let res = a[0].data.filter(v => a.slice(1).every(a => a.data.includes(v)));

console.log(res)

var a = [
 name: "Foo",
 id: "123",
 data: ["65d4ze", "65h8914d"]
 ,
 
 name: "Bar",
 id: "321",
 data: ["65d4ze", "894ver81"]
 
];

let res = a[0].data.filter(v => a.slice(1).every(a => a.data.includes(v)));

console.log(res)

share|improve this answer

edited Mar 25 at 12:28

Peter Mortensen

13.8k1987113

answered Mar 25 at 10:14

Pranav C BalanPranav C Balan

88.6k1391117

edited Mar 25 at 12:28

Peter Mortensen

13.8k1987113

edited Mar 25 at 12:28

Peter Mortensen

13.8k1987113

answered Mar 25 at 10:14

Pranav C BalanPranav C Balan

88.6k1391117

answered Mar 25 at 10:14

Pranav C BalanPranav C Balan

88.6k1391117

1

var a = [
 name: "Foo",
 id: "123",
 data: ["65d4ze", "65h8914d"]
 ,
 
 name: "Bar",
 id: "321",
 data: ["65d4ze", "894ver81"]
 
 ],
 b = ;
a.forEach(function(i) 
 i.data.forEach(function(j) 
 if (!b.hasOwnProperty(j)) 
 b[j] = 0;
 
 b[j] = b[j] + 1;
 );
);
c = []
for (var i in b) 
 if (b.hasOwnProperty(i)) 
 if (b[i] > 1) 
 c.push(i)
 
 

console.log(c);

share|improve this answer

answered Mar 25 at 10:12

Shree TiwariShree Tiwari

287112

add a comment | 

1

var a = [
 name: "Foo",
 id: "123",
 data: ["65d4ze", "65h8914d"]
 ,
 
 name: "Bar",
 id: "321",
 data: ["65d4ze", "894ver81"]
 
 ],
 b = ;
a.forEach(function(i) 
 i.data.forEach(function(j) 
 if (!b.hasOwnProperty(j)) 
 b[j] = 0;
 
 b[j] = b[j] + 1;
 );
);
c = []
for (var i in b) 
 if (b.hasOwnProperty(i)) 
 if (b[i] > 1) 
 c.push(i)
 
 

console.log(c);

share|improve this answer

answered Mar 25 at 10:12

Shree TiwariShree Tiwari

287112

add a comment | 

var a = [
 name: "Foo",
 id: "123",
 data: ["65d4ze", "65h8914d"]
 ,
 
 name: "Bar",
 id: "321",
 data: ["65d4ze", "894ver81"]
 
 ],
 b = ;
a.forEach(function(i) 
 i.data.forEach(function(j) 
 if (!b.hasOwnProperty(j)) 
 b[j] = 0;
 
 b[j] = b[j] + 1;
 );
);
c = []
for (var i in b) 
 if (b.hasOwnProperty(i)) 
 if (b[i] > 1) 
 c.push(i)
 
 

console.log(c);

share|improve this answer

answered Mar 25 at 10:12

Shree TiwariShree Tiwari

287112

var a = [
 name: "Foo",
 id: "123",
 data: ["65d4ze", "65h8914d"]
 ,
 
 name: "Bar",
 id: "321",
 data: ["65d4ze", "894ver81"]
 
 ],
 b = ;
a.forEach(function(i) 
 i.data.forEach(function(j) 
 if (!b.hasOwnProperty(j)) 
 b[j] = 0;
 
 b[j] = b[j] + 1;
 );
);
c = []
for (var i in b) 
 if (b.hasOwnProperty(i)) 
 if (b[i] > 1) 
 c.push(i)
 
 

console.log(c);

var a = [
 name: "Foo",
 id: "123",
 data: ["65d4ze", "65h8914d"]
 ,
 
 name: "Bar",
 id: "321",
 data: ["65d4ze", "894ver81"]
 
 ],
 b = ;
a.forEach(function(i) 
 i.data.forEach(function(j) 
 if (!b.hasOwnProperty(j)) 
 b[j] = 0;
 
 b[j] = b[j] + 1;
 );
);
c = []
for (var i in b) 
 if (b.hasOwnProperty(i)) 
 if (b[i] > 1) 
 c.push(i)
 
 

console.log(c);

var a = [
 name: "Foo",
 id: "123",
 data: ["65d4ze", "65h8914d"]
 ,
 
 name: "Bar",
 id: "321",
 data: ["65d4ze", "894ver81"]
 
 ],
 b = ;
a.forEach(function(i) 
 i.data.forEach(function(j) 
 if (!b.hasOwnProperty(j)) 
 b[j] = 0;
 
 b[j] = b[j] + 1;
 );
);
c = []
for (var i in b) 
 if (b.hasOwnProperty(i)) 
 if (b[i] > 1) 
 c.push(i)
 
 

console.log(c);

share|improve this answer

answered Mar 25 at 10:12

Shree TiwariShree Tiwari

287112

answered Mar 25 at 10:12

Shree TiwariShree Tiwari

287112

answered Mar 25 at 10:12

Shree TiwariShree Tiwari

287112

1

Use the flat function in the array:

var a = [
 name: "Foo",
 id: "123",
 data: ["65d4ze", "65h8914d"]
 ,
 
 name: "Bar",
 id: "321",
 data: ["65d4ze", "894ver81"]
 
 ],
 b = [],
 c = []
a.forEach(function(object) 
 b.push(object.data.map(function(val) 
 return val;
 )
 );
);

console.log(b.flat());

share|improve this answer

edited Mar 25 at 12:29

Peter Mortensen

13.8k1987113

answered Mar 25 at 10:18

thelastwormthelastworm

1428

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks for your answer, that a great code. But flat is not working on my browser
  
  – Kamoulox
  Mar 25 at 10:21
  

  
  
  
  

add a comment | 

1

Use the flat function in the array:

var a = [
 name: "Foo",
 id: "123",
 data: ["65d4ze", "65h8914d"]
 ,
 
 name: "Bar",
 id: "321",
 data: ["65d4ze", "894ver81"]
 
 ],
 b = [],
 c = []
a.forEach(function(object) 
 b.push(object.data.map(function(val) 
 return val;
 )
 );
);

console.log(b.flat());

share|improve this answer

edited Mar 25 at 12:29

Peter Mortensen

13.8k1987113

answered Mar 25 at 10:18

thelastwormthelastworm

1428

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks for your answer, that a great code. But flat is not working on my browser
  
  – Kamoulox
  Mar 25 at 10:21
  

  
  
  
  

add a comment | 

Use the flat function in the array:

var a = [
 name: "Foo",
 id: "123",
 data: ["65d4ze", "65h8914d"]
 ,
 
 name: "Bar",
 id: "321",
 data: ["65d4ze", "894ver81"]
 
 ],
 b = [],
 c = []
a.forEach(function(object) 
 b.push(object.data.map(function(val) 
 return val;
 )
 );
);

console.log(b.flat());

share|improve this answer

edited Mar 25 at 12:29

Peter Mortensen

13.8k1987113

answered Mar 25 at 10:18

thelastwormthelastworm

1428

var a = [
 name: "Foo",
 id: "123",
 data: ["65d4ze", "65h8914d"]
 ,
 
 name: "Bar",
 id: "321",
 data: ["65d4ze", "894ver81"]
 
 ],
 b = [],
 c = []
a.forEach(function(object) 
 b.push(object.data.map(function(val) 
 return val;
 )
 );
);

console.log(b.flat());

var a = [
 name: "Foo",
 id: "123",
 data: ["65d4ze", "65h8914d"]
 ,
 
 name: "Bar",
 id: "321",
 data: ["65d4ze", "894ver81"]
 
 ],
 b = [],
 c = []
a.forEach(function(object) 
 b.push(object.data.map(function(val) 
 return val;
 )
 );
);

console.log(b.flat());

var a = [
 name: "Foo",
 id: "123",
 data: ["65d4ze", "65h8914d"]
 ,
 
 name: "Bar",
 id: "321",
 data: ["65d4ze", "894ver81"]
 
 ],
 b = [],
 c = []
a.forEach(function(object) 
 b.push(object.data.map(function(val) 
 return val;
 )
 );
);

console.log(b.flat());

share|improve this answer

edited Mar 25 at 12:29

Peter Mortensen

13.8k1987113

answered Mar 25 at 10:18

thelastwormthelastworm

1428

edited Mar 25 at 12:29

Peter Mortensen

13.8k1987113

edited Mar 25 at 12:29

Peter Mortensen

13.8k1987113

answered Mar 25 at 10:18

thelastwormthelastworm

1428

answered Mar 25 at 10:18

thelastwormthelastworm

1428

1

You could use reduce and concat on each data array, and check the count of each item.

In the end, you check whether all objects across the array contain that item and return it if yes.

Note that this function works if you want to extract the item that has the same occurrence across all objects in the array.

If an item has duplicates, but does not fulfill the above condition, it would not be extracted.

let a = [name: "Foo",id: "123",data: ["65d4ze", "65h8914d"],name: "Bar",id: "321",data: ["65d4ze", "894ver81"]]

let arr = a.reduce((prev,next) => prev.data.concat(next.data))
let counts = ;
let result = [];
for (var i = 0; i < arr.length; i++) 
 var num = arr[i];
 counts[num] = counts[num] ? counts[num] + 1 : 1;


for (let i in counts) 
 if (counts[i] === a.length) 
 result.push(i)
 

console.log(result)

share|improve this answer

edited Mar 25 at 12:30

Peter Mortensen

13.8k1987113

answered Mar 25 at 10:55

tnkhtnkh

204110

add a comment | 

1

You could use reduce and concat on each data array, and check the count of each item.

In the end, you check whether all objects across the array contain that item and return it if yes.

Note that this function works if you want to extract the item that has the same occurrence across all objects in the array.

If an item has duplicates, but does not fulfill the above condition, it would not be extracted.

let a = [name: "Foo",id: "123",data: ["65d4ze", "65h8914d"],name: "Bar",id: "321",data: ["65d4ze", "894ver81"]]

let arr = a.reduce((prev,next) => prev.data.concat(next.data))
let counts = ;
let result = [];
for (var i = 0; i < arr.length; i++) 
 var num = arr[i];
 counts[num] = counts[num] ? counts[num] + 1 : 1;


for (let i in counts) 
 if (counts[i] === a.length) 
 result.push(i)
 

console.log(result)

share|improve this answer

edited Mar 25 at 12:30

Peter Mortensen

13.8k1987113

answered Mar 25 at 10:55

tnkhtnkh

204110

add a comment | 

You could use reduce and concat on each data array, and check the count of each item.

In the end, you check whether all objects across the array contain that item and return it if yes.

Note that this function works if you want to extract the item that has the same occurrence across all objects in the array.

If an item has duplicates, but does not fulfill the above condition, it would not be extracted.

let a = [name: "Foo",id: "123",data: ["65d4ze", "65h8914d"],name: "Bar",id: "321",data: ["65d4ze", "894ver81"]]

let arr = a.reduce((prev,next) => prev.data.concat(next.data))
let counts = ;
let result = [];
for (var i = 0; i < arr.length; i++) 
 var num = arr[i];
 counts[num] = counts[num] ? counts[num] + 1 : 1;


for (let i in counts) 
 if (counts[i] === a.length) 
 result.push(i)
 

console.log(result)

share|improve this answer

edited Mar 25 at 12:30

Peter Mortensen

13.8k1987113

answered Mar 25 at 10:55

tnkhtnkh

204110

let a = [name: "Foo",id: "123",data: ["65d4ze", "65h8914d"],name: "Bar",id: "321",data: ["65d4ze", "894ver81"]]

let arr = a.reduce((prev,next) => prev.data.concat(next.data))
let counts = ;
let result = [];
for (var i = 0; i < arr.length; i++) 
 var num = arr[i];
 counts[num] = counts[num] ? counts[num] + 1 : 1;


for (let i in counts) 
 if (counts[i] === a.length) 
 result.push(i)
 

console.log(result)

let a = [name: "Foo",id: "123",data: ["65d4ze", "65h8914d"],name: "Bar",id: "321",data: ["65d4ze", "894ver81"]]

let arr = a.reduce((prev,next) => prev.data.concat(next.data))
let counts = ;
let result = [];
for (var i = 0; i < arr.length; i++) 
 var num = arr[i];
 counts[num] = counts[num] ? counts[num] + 1 : 1;


for (let i in counts) 
 if (counts[i] === a.length) 
 result.push(i)
 

console.log(result)

let a = [name: "Foo",id: "123",data: ["65d4ze", "65h8914d"],name: "Bar",id: "321",data: ["65d4ze", "894ver81"]]

let arr = a.reduce((prev,next) => prev.data.concat(next.data))
let counts = ;
let result = [];
for (var i = 0; i < arr.length; i++) 
 var num = arr[i];
 counts[num] = counts[num] ? counts[num] + 1 : 1;


for (let i in counts) 
 if (counts[i] === a.length) 
 result.push(i)
 

console.log(result)

share|improve this answer

edited Mar 25 at 12:30

Peter Mortensen

13.8k1987113

answered Mar 25 at 10:55

tnkhtnkh

204110

edited Mar 25 at 12:30

Peter Mortensen

13.8k1987113

edited Mar 25 at 12:30

Peter Mortensen

13.8k1987113