jquery index() returning -1

Question

Why does my imageIndex keep returning -1;

    $(function(){
            //rotateImages();
        setInterval("rotateImages()", 4000);
    });

    var imageIndex = 0;
    var currentPhoto = $("#photoShow div.current");

    function rotateImages(){
        var max = $("#photoShow div").length;
        imageIndex = currentPhoto.index();
        console.log(imageIndex + "    ::   "+ (max - 1));

    }

HTML :

<body>

<div id="photoShow">
    <div class="current">
        <img src="images/Grass.jpg" alt="Photo Gallery" width="400" height="400" class="gallery" />
    </div>
    <div>
        <img src="images/Leaf.jpg" alt="Photo Gallery" width="400" height="400" class="gallery" />
    </div>
    <div>
        <img src="images/Spring.jpg" alt="Photo Gallery" width="400" height="400" class="gallery" />
    </div>
    <div>
        <img src="images/Water.jpg" alt="Photo Gallery" width="400" height="400" class="gallery" />
    </div>
</div>

Returns 0 for me, which is correct.

Marcel
– Marcel

2011-05-20 04:30:47 +00:00
Commented May 20, 2011 at 4:30 — Marcel
– Marcel, Commented May 20, 2011 at 4:30

Daniel · Accepted Answer · 2011-05-20 04:46:15Z

3

$.index() returns -1 if the element is not found.

Based on your source code above, it looks like you're not waiting until the DOM is ready before declaring currentPhoto...so chances are the element doesn't exist yet in the DOM when the variable is being defined.

Simply moving everything into $(function(){...}) would give you the results you're expecting.

$(function(){
    //rotateImages();
    setInterval("rotateImages()", 4000);
    var imageIndex = 0;
    var currentPhoto = $("#photoShow div.current");

    function rotateImages(){
        var max = $("#photoShow div").length;
        imageIndex = currentPhoto.index();
        console.log(imageIndex + "    ::   "+ (max - 1));
    }
});

answered May 20, 2011 at 4:46

Daniel

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1 Comment

Chapsterj Over a year ago

When I moved it all inside the initial jquery function the setInterval can no longer find the function rotateImages(). I get an error:: ReferanceError: Can't find variable: rotateImages

dkamins · Accepted Answer · 2011-05-20 04:25:52Z

1

Try:

imageIndex = $('#photoShow').index('.current');

answered May 20, 2011 at 4:25

dkamins

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Comments

Mayank · Accepted Answer · 2011-05-20 04:34:47Z

0

This statement var currentPhoto = $("#photoShow div.current"); has length 1 as your code has only one <div class="current">

imageIndex = currentPhoto.index(); should return 0 in that case. In console logs 0 - 1 = -1 is the output

EDIT:
Sorry, your are doing var max = $("#photoShow div").length; and max - 1. imageIndex should be zero :-(

edited May 20, 2011 at 4:34

answered May 20, 2011 at 4:27

Mayank

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