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

Question

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.)

$shapes.length is undefined... so.. then it's not actually an array, even though I assign values to it like $shapes[id] = val; so... how do I search through it then? — NullVoxPopuli
– NullVoxPopuli, Commented Sep 26, 2011 at 14:52
Usually, when questions are of the form 'Why doesn't <insert name of popular library function> work?' come along - it is being used incorrectly. $shapes may not be an Array, also comparison of objects is prone to trouble. — Orbling
– Orbling, Commented Sep 26, 2011 at 14:53
@TheLindyHop, using jQuery.each(), but, have a mind to how you compare elements. — Orbling
– Orbling, Commented Sep 26, 2011 at 14:58

SLaks · Accepted Answer · 2011-09-26 14:53:44Z

3

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!
    }
}

answered Sep 26, 2011 at 14:53

SLaks

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2 Comments

NullVoxPopuli Over a year ago

how? I've tried .each, for loops... not really sure what this object is either.

NullVoxPopuli Over a year ago

nice, I found a similar solution, added to OP

Niko · Accepted Answer · 2011-09-26 14:53:59Z

3

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.

answered Sep 26, 2011 at 14:53

Niko

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