jquery onmouseover function

Question

i need to rotate an image onmouseover function using Jquery

file.js

$("#logo").rotate({
        bind:
         {
             mouseover: function () 
             {$(this).rotate({ animateTo: 180 })},
             mouseout: function () 
             {$(this).rotate({ animateTo: 0 })}
        }
 });

HTML

<script type="text/javascript" src="../script/file.js"></script>
<img id="logo" src="../images/logo.png" alt="none" />

Is there anything wrong in this code? BTW, I found this code on the internet. There is no function call as rotate() in JavaScript then how can it work? Please give any link so that I can understand.

Please ignore this question if it looks like a foolish thing to ask :)

http://stackoverflow.com/questions/7936396/jquery-rotate-transform — disperse, Aug 29 '12 at 16:34

score 1 · Answer 1 · answered Aug 29 '12 at 16:20

1

.rotate() isn't a function. Are you referencing whatever plugin this is correctly? This must be what's causing the issue.

answered Aug 29 '12 at 16:20

Mark Pieszak - Trilon.io

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i need to call the function from image onmouseover() what should i replace with rotate() then? – Mr_Green Aug 29 '12 at 16:21

score 0 · Answer 2 · answered Aug 29 '12 at 16:22

It looks like you are attempting to use some code snippet that is using a custom JQuery plugin.

There is no built-in function for rotating images in JQuery. There are plenty of plugins available for that, e.g. http://code.google.com/p/jquery-rotate/ and http://code.google.com/p/jqueryrotate/.

score 0 · Accepted Answer · answered Aug 29 '12 at 16:39

0

Download jquery.rotate.1-1.js from here http://code.google.com/p/jquery-rotate/downloads/list and include it in a script tag before this call.

answered Aug 29 '12 at 16:39

Victor 'Chris' Cabral

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jquery onmouseover function

3 Answers3