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Question

Answer 1

+4 A:

You need to put it in quotes - it's a string:

document.getElementById('loadingImage').style.visibility='visible';

David M 2010-01-04 11:54:07

Answer 2

A:

I would use Jquery. Go download it at the Jquery home page.

Then, include it:

<script type="text/javascript" src="http://jqueryjs.googlecode.com/files/jquery-1.3.2.min.js"&gt;&lt;/script&gt;
<script type="text/javascript">
function showImage(){
$("#loadingImage").toggle();
}

</script>


<img id="loadingImage" src="ajax-loader.gif" style="display:none;"/>

TIMEX 2010-01-04 11:54:46

Answer 3

A:

If the other answers don't give you the results you're after, try setting the display to none:

document.getElementById('loadingImage').style.display='none';

Phil.Wheeler 2010-01-04 11:55:34

Answer 4

A:

You need to enclose it in quote marks, otherwise JavaScript thinks you're trying to set it to be the value of a variable called "visible". Since you don't have a variable called "visible", you get the error saying that it's undefined.

document.getElementById('loadingImage').style.visibility='visible';

Rich Adams 2010-01-04 11:55:40

Answer 5

A:

I am very very sorry. It should be

  document.getElementById('loadingImage').style.visibility='visible';

quuotes missing aroung visible.

abhishek 2010-01-04 11:56:20

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show/hide image on click

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