如何检查Javascript数组中是否存在多个值

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.

For instance, example 1...

Array A contains the following values

34, 78, 89

Array B contains the following values

78, 67, 34, 99, 56, 89

This would return true

...example 2:

Array A contains the following values

34, 78, 89

Array B contains the following values

78, 67, 99, 56, 89

This would return false

...example 3:

Array A contains the following values

34, 78, 89

Array B contains the following values

78, 89

This would return false

So far I have tried to solve this by:

1. Extending Jquery with a custom 'compare' method to compare the two arrays. Problem is this only returns true when the arrays are identical and as you can see from example 1 I want it to return true even if they aren't identical but at least contain the value
2. using Jquerys .inArray function , but this only checks for one value in an array, not multiple.

Any light that anyone could throw on this would be great.

Answer 1

Native JavaScript solution

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

---

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() .

---

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 .

---

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

A one-liner to test that all of the elements in arr1 exist in arr2 ...

With es6:

var containsAll = arr1.every(i => arr2.includes(i));

Without es6:

var containsAll = arr1.every(function (i) { return arr2.includes(i); });

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.

Answer 5

You could take a Set and check all items agains it.

 const containsAll = (needles, haystack) => needles.every(Set.prototype.has, new Set(haystack)); console.log(containsAll([105, 112, 103], [106, 105, 103, 112]));

Answer 6

If you need a little bit more visibility on which items are in the array you can use this one :

var tools = {
        elem : {},
        arrayContains : function(needles, arrhaystack) {
           if (this.typeOf(needles) === 'array') {
                needle.reduce(function(result,item,$i,array){ // You can use any other way right there.
                    var present = (arrhaystack.indexOf(item) > -1);
                    Object.defineProperty(tools.elem, item, {
                        value : present,
                        writable : true
                    });
                },{})
                return this.elem;
            }
        },        
        typeOf : function(obj) {
            return {}.toString.call(obj).split(' ')[1].slice(0, -1).toLowerCase();
        }
    }

Use it with simply var check = tools.arrayContains([10,'foo'], [1,'foo','bar'])

Then you get the result like

10 : false
foo : true

Then if you need to get only one result if one of them is true you can :

arr = Object.values(check);
(arr.indexOf('true')) ? instru1 : instru2 ;

I don't think that's the better way but it's working & easily adaptable. Considering this example I advise you to make an Object.create(tools) before use it in your way.

Answer 7

Using array functions: [].filter and [].includes

Something like this:

[34, 78, 89].filter((v) => {
    return [78, 67, 34, 99, 56, 89].includes(v);
});

This will return an array of the matches items

Then we can compare it with needles array

As a function it will be:

const contains = (haystack, needles) => {
    return haystack.filter((v) => {
        return needles.includes(v);
    }).length === needles.length;
}

Answer 8

Just for the fun of it, i've implemented something with Array.prototype.reduce() :

let areKeysPresent = function (sourceArray, referenceArray) {
    return sourceArray.reduce((acc, current) => acc & referenceArray.includes(current), true)
}

Answer 9

You could also add a containsAll to Array prototype like this:

Array.prototype.containsAll = function()
{
    return Array.from(arguments).every(i => this.includes(i));
}

Examples:

["apple", "banana", "strawberry"].containsAll("apple", "banana"); // true

["apple", "banana", "strawberry"].containsAll("apple", "kiwi"); // false

Answer 10

You can use this simple function (renamed variables as per above answer for easy reading):

function contains(haystack, needles) {

    return needles.map(function (needle) { 
        return haystack.indexOf(needle);
    }).indexOf(-1) == -1;
}

Answer 11

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 
}