Best way to find if an item is in a JavaScript array?

If the array is unsorted, there isn't really a better way (aside from using the above-mentioned indexOf, which I think amounts to the same thing). If the array is sorted, you can do a binary search, which works like this:

Pick the middle element of the array.
Is the element you're looking for bigger than the element you picked? If so, you've eliminated the bottom half of the array. If it isn't, you've eliminated the top half.
Pick the middle element of the remaining half of the array, and continue as in step 2, eliminating halves of the remaining array. Eventually you'll either find your element or have no array left to look through.
Binary search runs in time proportional to the logarithm of the length of the array, so it can be much faster than looking at each individual element.

4 years ago

If you are using jQuery:

$.inArray(5 + 5, [ "8", "9", "10", 10 + "" ]);

4 years ago

Assuming .indexOf() is implemented

Object.defineProperty( Array.prototype,'has',

{

    value:function(o, flag){

    if (flag === undefined) {

        return this.indexOf(o) !== -1;

    } else {   // only for raw js object

        for(var v in this) {

            if( JSON.stringify(this[v]) === JSON.stringify(o)) return true;

        }

        return false;                       

    },

    // writable:false,

    // enumerable:false

})

!!! do not make

Array.prototype.has=function(){...

because you'll add an enumerable element in every array and js is broken.

//use like          

[22 ,'a', {prop:'x'}].has(12) // false

["a","b"].has("a") //  true



[1,{a:1}].has({a:1},1) // true

[1,{a:1}].has({a:1}) // false

the use of 2nd arg (flag) forces comparison by value instead of reference

comparing raw objects

[o1].has(o2,true) // true if every level value is same

4 years ago