How to check whether multiple values exist within an Javascript array

Question

So, I\'m using Jquery and have two arrays both with multiple values and I want to check whether all the values in the first array exist in the second.

Answer 1

Native JavaScript solution

var success = array_a.every(function(val) {
    return array_b.indexOf(val) !== -1;
});

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You'll need compatibility patches for every and indexOf if you're supporting older browsers, including IE8.

• Compatibility patch from MDN for .every().
• Compatibility patch from MDN for .indexOf().

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Full jQuery solution

var success = $.grep(array_a, function(v,i) {
    return $.inArray(v, array_b) !== -1;
}).length === array_a.length;

Uses $.grep with $.inArray.

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ES2015 Solution

The native solution above can be shortened using ES2015's arrow function syntax and its .includes() method:

let success = array_a.every((val) => array_b.includes(val))

Answer 2

function containsAll(needles, haystack){ 
  for(var i = 0; i < needles.length; i++){
     if($.inArray(needles[i], haystack) == -1) return false;
  }
  return true;
}

containsAll([34, 78, 89], [78, 67, 34, 99, 56, 89]); // true
containsAll([34, 78, 89], [78, 67, 99, 56, 89]); // false
containsAll([34, 78, 89], [78, 89]); // false

Answer 3

Try this.

var arr1 = [34, 78, 89];
var arr2 = [78, 67, 34, 99, 56, 89];

var containsVal = true;
$.each(arr1, function(i, val){
   if(!$.inArray(val, arr2) != -1){
       retVal = false;
       return false;
   }
});

if(containsVal){
    //arr2 contains all the values from arr1 
}

Answer 4

I noticed that the question is about solving this with jQuery, but if anyone else who is not limited to jQuery comes around then there is a simple solution using underscore js.

Using underscore js you can do:

_.intersection(ArrayA, ArrayB).length === ArrayA.length;

From the docs:

intersection_.intersection(*arrays) Computes the list of values that are the intersection of all the arrays. Each value in the result is present in each of the arrays.

_.intersection([1, 2, 3], [101, 2, 1, 10], [2, 1]); => [1, 2]

Ergo, if one of the items in ArrayA was missing in ArrayB, then the intersection would be shorter than ArrayA.