is the anyway to use $.inArray (or alternative) with a javascript array of objects?

Question

I have the following array in javascript:

var myFirstArray = [1,2,3];

and i use $.inArrray() to see if a specific number is in that array like this

 var num =3;
 var exists = $.inArray(num, myFirstArray) > -1;

I now have an array of objects

var myArray = [{value:1, label:"bird"}, {value:2, label:"dog"}, {value:3, label: "cat"}];

if there anyway to use $.inArray() to search within a field of an object? Something like this:

var num = 3;
var exists = $.inArray(num, myFirstArray, (r) {return r.value}) > -1;

if the answer is NO, is there an alternative function that would give me this behavior that is performant?

No. See the documentation: inArray does not allow a predicate function. — user2864740
– user2864740, Commented Nov 23, 2013 at 20:27

Matyas Matyas · Accepted Answer · 2013-11-23 21:05:14Z

2

Alternatively you can use one of the jQuery methods grep or each.

var num = 3;
var myArray = [{value: 1, label: "bird"}, {value:2, label:"dog"}, {value:3, label: "cat"}];

var result = jQuery.grep(myArray, function( element, index ) {
    return ( num === element.value );
});

var found = false;
$.each(myArray, function ( index, element ) {
    if ( num === element.value ) {
        found = true;
        return false;
   }
});

edited Nov 23, 2013 at 21:05

answered Nov 23, 2013 at 20:52

Matyas Matyas

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Fabrício Matté · Accepted Answer · 2013-11-23 20:41:26Z

2

You can use the native Array#some method:

var num =3;
var myArray = [{value:1, label:"bird"}, {value:2, label:"dog"}, {value:3, label: "cat"}];

var exists = myArray.some(function(o) {
    return o.value === num;
});

Live demo

If you need support for pesky IE<9: MDN Array.prototype.some shim or you may also use a complete ES5 shim already (thanks @user2864740). Underscore also provides the _.some utility.

edited Nov 23, 2013 at 20:41

answered Nov 23, 2013 at 20:35

Fabrício Matté

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1 Comment

Fabrício Matté Over a year ago

@user2864740 Thanks, added the link to the answer. =]

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