Javascript Checking array for presence of a specific number

依然范特西╮ 提交于 2019-12-01 20:45:49

What you can do is write yourself a function to check if an element belongs to an array:

function inArray(array, value) {
    for (var i = 0; i < array.length; i++) {
        if (array[i] == value) return true;
    }
    return false;
}

And the just do:

var arr = ["1","8","17","14","11","20","2","6"];
if (inArray(arr, 8)) {
    // change picture.src to srcpicture1
} else {
    // change picture.src to srcpicture2
}

It's a lot more readable to me.


For extra points you can add the function to the array prototype like so:

Array.prototype.has = function (value) {
    for (var i = 0; i < this.length; i++) {
        if (this[i] === value) return true;
    }
    return false;
};

And then the call would be

if (arr.has(8)) // ...

Pushing this even further, you can check for indexOf() method on array and use it - if not - replace it with the code above.


P.S. Try not to use Array for a variable name, since it's reserved for the actual array type.

Why don't just you abort the loop when you find the right number :

for(x=0;x<=Array.length;x++)
{
    if(Array[x]==8) {
        //change picture.src to srcpicture1
        break;
    }
}

You could sort the array first then check the array only up to the point at which a number would be in the array, were it to exist.

If you have unique keys and a faster retrieval is what you care about a lot, you can consider using a map instead of an array (if there's a hard-bound case of using an array, then it won't work of course). If using a map, you just check "if( num in arr ) ".

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