CODEWITHSUNDEEP
javascriptjqueryarrays
NullVoxPopuli
NullVoxPopuli

Reputation: 65143

jQuery: why doesn't jQuery.inArray() work?

enter image description here

As shown in the image, I just do inArray on an array, looking for a node. and $previously_selected_node and the item at index 37 in $shapes are the same object.... so... why isn't it working?

EDIT: I found another way to search after one of the aswerers postedd his answer:

var result = -1;
jQuery.each(shapes, function(key, value){
    if (value.id == shape.id){
        result = key;
    }
});
return result;

apparently, part of my problem is that I can't return in the middle of a loop. (I was returning the instant a match was found, which was causing some issues.)

Upvotes: 3

Views: 1544

Answers (2)

Niko
Niko

Reputation: 26730

You're doing it the wrong way round. It's actually $.inArray(value, array).

And as others already stated: inArray is for arrays, not for objects.

Upvotes: 3

SLaks
SLaks

Reputation: 887453

Your object is not an array.
$.inArray only work on array-like objects with a length and a set of properties named 0 through length - 1.

You need to search your non-array manually.
For example, you could use a for / in loop to loop through all properties that actually exist and see if any of them match your object:

for (var key in $shapes) {
    if ($shapes[key] === yourObject) {
        //Match!
    }
}

Upvotes: 3

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