jquery version of array.contains

Question

Can jQuery test an array for the presence of an object (either as part of the core functionality or via an avaible plugin)?

Also, I\'m looking for something like a

Answer 1

I found way to remove object:

foot = { bar : 'test'};
delete foot[bar];

Answer 2

This is not jQuery, but in one line you can add a handy 'contains' method to arrays. I find this helps with readability (especially for python folk).

Array.prototype.contains = function(a){ return this.indexOf(a) != -1 }

example usage

 > var a = [1,2,3]
 > a.contains(1)
true
 > a.contains(4)
false

Similarly for remove

Array.prototype.remove = function(a){if (this.contains(a)){ this.splice(this.indexOf(a),1)}; return this}

> var a = [1,2,3]
> a.remove(2)
[1,3]

Or, if you want it to return the thing removed rather than the altered array, then

Array.prototype.remove = function(a){if (this.contains(a)){ return this.splice(this.indexOf(a),1)}}

> var a = [1,2,3]
> a.remove(2)
[2]
> a
[1,3]

Answer 3

If your list contains a list of elements, then you can use jQuery.not or jQuery.filter to do your "array.remove". (Answer added because of the high google score of your original question).

Answer 4

jQuery.inArray returns the first index that matches the item you searched for or -1 if it is not found:

if($.inArray(valueToMatch, theArray) > -1) alert("it's in there");

You shouldn't need an array.remove. Use splice:

theArray.splice(startRemovingAtThisIndex, numberOfItemsToRemove);

Or, you can perform a "remove" using the jQuery.grep util:

var valueToRemove = 'someval';
theArray = $.grep(theArray, function(val) { return val != valueToRemove; });