Change DIV Background on rollover with Jquery

Question

I am very new to JQuery but already like it a lot.

I am trying to change a background of a div when the user rollsovers it. But I can not get the bg to update. So far this is what I have:

    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
<style>
#fish{background-image:url(images/fish_off.jpg);width:459px;height:474px;}
</style>

 <script type="text/javascript" src="javascript/jquery.js"></script>     

 <script type="text/javascript">                                         
$('#fish').hover(
  function(){$('#fish').css({background: "url(images/fish_on.jpg)"})},
  function(){$('#fish').css({background: "url(images/fish_off.jpg)"})}
);


 </script>      

</head>

<body>

<div id="fish"></div>

</body>
</html>

Here is my test page and based off the info below I am still having issues:

Test Page

Community · Accepted Answer · 2017-05-23 12:08:54Z

5

use .mouseenter() or if you want to apply a class on mouseout as well use .hover() http://api.jquery.com/mouseenter/ http://api.jquery.com/hover/

$("div").hover(
  function () {
      $(this).css("background", "#ddd");
  }, 
  function () {
      $(this).css("background", "#ccc");
  }
);

you can also do this purely through css if you just want a rollver

    #fish { background: #fff; 
    #fish:hover { background: #ccc; }

check out this working fiddle on using hover http://jsfiddle.net/faLdY/

Applying background images via jquery is slightly different checkout this similar question Switching a DIV background image with jQuery

note the way the css is applied in your example page you are doing it incorrectly.

Change

$('#fish').hover(
  function(){$('#fish').css({background: "url(images/fish_on.jpg)"})},
  function(){$('#fish').css({background: "url(images/fish_off.jpg)"})}
);

to this

$('#fish').hover(
  function(){$('#fish').css("background-image", "url(images/fish_on.jpg)")},
  function(){$('#fish').css("background-image", "url(images/fish_off.jpg)")}
);

EDIT:

Here is a working fiddle doing exactly what you asked http://jsfiddle.net/C7KxR/

this is linked directly to your images, btw. I hope thats alright.

edited May 23, 2017 at 12:08

CommunityBot

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answered May 13, 2011 at 1:52

matchew

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7 Comments

Denoteone Over a year ago

I would normally use css but I am also going to show a hidden div figured I would do them both at the same time

matchew Over a year ago

fair enough. If you want to show a hidden div perhaps look into toggle, or slideToggle() or fadeToggle() and also the jquery-ui library if you want more control over the animation.

Denoteone Over a year ago

I applied your feedback but am still not seeing the change. I added a link above to the page in my post. Not sure what I am doing wrong?

matchew Over a year ago

@Denoteone I think your problem is with the way you are applying the .css, I have updated my answer after reviewing your test page.

Denoteone Over a year ago

Thank you for all your help. Could the issue have to do with my reference to the jquery library?

|

Yisroel · Accepted Answer · 2011-05-13 01:53:05Z

1

change

$('#fish').click(function()
{
  $('#fish').css("background-image", "url(images/fish_on.jpg)");  
});

to

$('#fish').hover(function()
{
  $('#fish').css("background-image", "url(images/fish_on.jpg)");  
}, function()
{
  $('#fish').css("background-image", "url(images/fish_ooff.jpg)");  
});

answered May 13, 2011 at 1:53

Yisroel

8,1744 gold badges29 silver badges26 bronze badges

3 Comments

Denoteone Over a year ago

Just so I under stand the second function() is to swap the image back to normal...correct?

Yisroel Over a year ago

Yes. .hover() takes 2 functions, one on mouseenter, the second on mouseleave. see api.jquery.com/hover

wewals Over a year ago

Yes, that's correct. Hover is just mainly an alias for mouseenter and mouseleave.

wewals · Accepted Answer · 2011-05-13 01:55:08Z

1

$('#fish').hover(
  function(){$('#fish').css({background: "url(images/fish_on.jpg)"})},
  function(){$('#fish').css({background: "url(images/fish_off.jpg)"})}
);

You want to use .hover which takes two callbacks. One for mousein and one for mouseout.

answered May 13, 2011 at 1:55

wewals

1,4479 silver badges9 bronze badges

Comments

Sergei · Accepted Answer · 2011-05-13 01:52:51Z

0

you need mouseover event:

$('#fish').mouseover(...

answered May 13, 2011 at 1:52

Sergei

2,7571 gold badge17 silver badges17 bronze badges

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