43

In JavaScript, how do I test that one array has the elements of another array?

arr1 = [1, 2, 3, 4, 5]
[8, 1, 10, 2, 3, 4, 5, 9].function_name(arr1) # => true
Zeck
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  • Does this answer your question? [Check if array contains all elements of another array](https://stackoverflow.com/questions/53606337/check-if-array-contains-all-elements-of-another-array) – Michael Freidgeim Nov 19 '21 at 11:52

4 Answers4

83

No set function does this, but you can simply do an ad-hoc array intersection and check the length.

[8, 1, 10, 2, 3, 4, 5, 9].filter(function (elem) {
    return arr1.indexOf(elem) > -1;
}).length == arr1.length

A more efficient way to do this would be to use .every which will short circuit in falsy cases.

arr1.every(elem => arr2.indexOf(elem) > -1);
Explosion Pills
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22

You can use array.indexOf():

pseudocode:

function arrayContainsAnotherArray(needle, haystack){
  for(var i = 0; i < needle.length; i++){
    if(haystack.indexOf(needle[i]) === -1)
       return false;
  }
  return true;
}
valentinas
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4

ES6 solution using includes:

[1].every(elem => [1,2,3].includes(elem));

Very similar to Explosion Pills's solution above, just a bit more readable (and arguably a tiny, tiny bit slower).

Martin
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3
function arr(arr1,arr2)
{
    for(var i=0;i<arr1.length;i++)
     {
        if($.inArray(arr1[i],arr2) ==-1)
               //here it returns that arr1 value does not contain the arr2
        else
             // here it returns that arr1 value contains in arr2

     }

}
tamilmani
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