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JavaScript array of objects contains the same array data


Simplest code for array intersection in javascriptHow do JavaScript closures work?What is the most efficient way to deep clone an object in JavaScript?How do I remove a property from a JavaScript object?How do I check if an array includes an object in JavaScript?What does “use strict” do in JavaScript, and what is the reasoning behind it?How to check whether a string contains a substring in JavaScript?Loop through an array in JavaScriptHow to check if an object is an array?How do I remove a particular element from an array in JavaScript?For-each over an array in JavaScript?













7















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);












share|improve this question









New contributor




Kamoulox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • So get all arrays elements that are present on all a?

    – Eddie
    yesterday






  • 3





    Possible duplicate of Simplest code for array intersection in javascript

    – James Long
    yesterday











  • @JamesLong I don't understand the duplication, why this answer can help me ?

    – Kamoulox
    yesterday











  • @Eddie I need to get all array into the key data. This key is present into all object of my principal array.

    – Kamoulox
    yesterday











  • @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
    yesterday















7















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);












share|improve this question









New contributor




Kamoulox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • So get all arrays elements that are present on all a?

    – Eddie
    yesterday






  • 3





    Possible duplicate of Simplest code for array intersection in javascript

    – James Long
    yesterday











  • @JamesLong I don't understand the duplication, why this answer can help me ?

    – Kamoulox
    yesterday











  • @Eddie I need to get all array into the key data. This key is present into all object of my principal array.

    – Kamoulox
    yesterday











  • @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
    yesterday













7












7








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);












share|improve this question









New contributor




Kamoulox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












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);








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);






javascript arrays






share|improve this question









New contributor




Kamoulox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




Kamoulox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited yesterday









Peter Mortensen

13.8k1987113




13.8k1987113






New contributor




Kamoulox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked yesterday









KamouloxKamoulox

413




413




New contributor




Kamoulox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Kamoulox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Kamoulox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • So get all arrays elements that are present on all a?

    – Eddie
    yesterday






  • 3





    Possible duplicate of Simplest code for array intersection in javascript

    – James Long
    yesterday











  • @JamesLong I don't understand the duplication, why this answer can help me ?

    – Kamoulox
    yesterday











  • @Eddie I need to get all array into the key data. This key is present into all object of my principal array.

    – Kamoulox
    yesterday











  • @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
    yesterday

















  • So get all arrays elements that are present on all a?

    – Eddie
    yesterday






  • 3





    Possible duplicate of Simplest code for array intersection in javascript

    – James Long
    yesterday











  • @JamesLong I don't understand the duplication, why this answer can help me ?

    – Kamoulox
    yesterday











  • @Eddie I need to get all array into the key data. This key is present into all object of my principal array.

    – Kamoulox
    yesterday











  • @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
    yesterday
















So get all arrays elements that are present on all a?

– Eddie
yesterday





So get all arrays elements that are present on all a?

– Eddie
yesterday




3




3





Possible duplicate of Simplest code for array intersection in javascript

– James Long
yesterday





Possible duplicate of Simplest code for array intersection in javascript

– James Long
yesterday













@JamesLong I don't understand the duplication, why this answer can help me ?

– Kamoulox
yesterday





@JamesLong I don't understand the duplication, why this answer can help me ?

– Kamoulox
yesterday













@Eddie I need to get all array into the key data. This key is present into all object of my principal array.

– Kamoulox
yesterday





@Eddie I need to get all array into the key data. This key is present into all object of my principal array.

– Kamoulox
yesterday













@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
yesterday





@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
yesterday












5 Answers
5






active

oldest

votes


















11














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























  • 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
    yesterday






  • 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
    yesterday











  • With array common = array .map(o => o[key]) .reduce((a, b) => b.filter(Array.prototype.includes, a));

    – Pranav C Balan
    yesterday











  • @PranavCBalan, Array#includes uses a second parameter fromIndex and gets this value with the index of the callback ...

    – Nina Scholz
    yesterday












  • @NinaScholz : oops that's right, that would make issues :)

    – Pranav C Balan
    yesterday



















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
































    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






























      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

























      • Thanks for your answer, that a great code. But flat is not working on my browser

        – Kamoulox
        yesterday


















      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
























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        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        11














        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























        • 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
          yesterday






        • 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
          yesterday











        • With array common = array .map(o => o[key]) .reduce((a, b) => b.filter(Array.prototype.includes, a));

          – Pranav C Balan
          yesterday











        • @PranavCBalan, Array#includes uses a second parameter fromIndex and gets this value with the index of the callback ...

          – Nina Scholz
          yesterday












        • @NinaScholz : oops that's right, that would make issues :)

          – Pranav C Balan
          yesterday
















        11














        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























        • 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
          yesterday






        • 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
          yesterday











        • With array common = array .map(o => o[key]) .reduce((a, b) => b.filter(Array.prototype.includes, a));

          – Pranav C Balan
          yesterday











        • @PranavCBalan, Array#includes uses a second parameter fromIndex and gets this value with the index of the callback ...

          – Nina Scholz
          yesterday












        • @NinaScholz : oops that's right, that would make issues :)

          – Pranav C Balan
          yesterday














        11












        11








        11







        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













        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);








        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












        share|improve this answer



        share|improve this answer










        answered yesterday









        Nina ScholzNina Scholz

        193k15107178




        193k15107178












        • 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
          yesterday






        • 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
          yesterday











        • With array common = array .map(o => o[key]) .reduce((a, b) => b.filter(Array.prototype.includes, a));

          – Pranav C Balan
          yesterday











        • @PranavCBalan, Array#includes uses a second parameter fromIndex and gets this value with the index of the callback ...

          – Nina Scholz
          yesterday












        • @NinaScholz : oops that's right, that would make issues :)

          – Pranav C Balan
          yesterday


















        • 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
          yesterday






        • 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
          yesterday











        • With array common = array .map(o => o[key]) .reduce((a, b) => b.filter(Array.prototype.includes, a));

          – Pranav C Balan
          yesterday











        • @PranavCBalan, Array#includes uses a second parameter fromIndex and gets this value with the index of the callback ...

          – Nina Scholz
          yesterday












        • @NinaScholz : oops that's right, that would make issues :)

          – Pranav C Balan
          yesterday

















        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
        yesterday





        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
        yesterday




        5




        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
        yesterday





        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
        yesterday













        With array common = array .map(o => o[key]) .reduce((a, b) => b.filter(Array.prototype.includes, a));

        – Pranav C Balan
        yesterday





        With array common = array .map(o => o[key]) .reduce((a, b) => b.filter(Array.prototype.includes, a));

        – Pranav C Balan
        yesterday













        @PranavCBalan, Array#includes uses a second parameter fromIndex and gets this value with the index of the callback ...

        – Nina Scholz
        yesterday






        @PranavCBalan, Array#includes uses a second parameter fromIndex and gets this value with the index of the callback ...

        – Nina Scholz
        yesterday














        @NinaScholz : oops that's right, that would make issues :)

        – Pranav C Balan
        yesterday






        @NinaScholz : oops that's right, that would make issues :)

        – Pranav C Balan
        yesterday














        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





























          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



























            3












            3








            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















            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)








            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














            share|improve this answer



            share|improve this answer








            edited yesterday









            Peter Mortensen

            13.8k1987113




            13.8k1987113










            answered yesterday









            Pranav C BalanPranav C Balan

            87.7k1391117




            87.7k1391117





















                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



























                  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

























                    1












                    1








                    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
















                    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












                    share|improve this answer



                    share|improve this answer










                    answered yesterday









                    Shree TiwariShree Tiwari

                    287112




                    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

























                        • Thanks for your answer, that a great code. But flat is not working on my browser

                          – Kamoulox
                          yesterday















                        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

























                        • Thanks for your answer, that a great code. But flat is not working on my browser

                          – Kamoulox
                          yesterday













                        1












                        1








                        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















                        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());








                        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














                        share|improve this answer



                        share|improve this answer








                        edited yesterday









                        Peter Mortensen

                        13.8k1987113




                        13.8k1987113










                        answered yesterday









                        thelastwormthelastworm

                        1428




                        1428












                        • Thanks for your answer, that a great code. But flat is not working on my browser

                          – Kamoulox
                          yesterday

















                        • Thanks for your answer, that a great code. But flat is not working on my browser

                          – Kamoulox
                          yesterday
















                        Thanks for your answer, that a great code. But flat is not working on my browser

                        – Kamoulox
                        yesterday





                        Thanks for your answer, that a great code. But flat is not working on my browser

                        – Kamoulox
                        yesterday











                        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





























                          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



























                            1












                            1








                            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















                            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)








                            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














                            share|improve this answer



                            share|improve this answer








                            edited yesterday









                            Peter Mortensen

                            13.8k1987113




                            13.8k1987113










                            answered yesterday









                            tnkhtnkh

                            17210




                            17210




















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                                Kamoulox is a new contributor. Be nice, and check out our Code of Conduct.












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                                Kamoulox is a new contributor. Be nice, and check out our Code of Conduct.














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