As for a proper example i can't do that too well as i don't have any
original code to work with but something like this may work:
html:
<div id="container">
<img src="path/to/img.ext"/>
<div id="menu" class="menu">
<!-- menu stuff here -->
</div>
</div>
JScript:
<script>
//Add a listener to the #container div, rather than the img.
$('#container').mouseover(function(){ $('#menu').show() });
$('#container').mouseout(function(){ $('#menu').hide() });
</script>
Without any example from you, this may well be useless unfortunately,
but the theory, at least to me, should work.
HTH
Mark
On Sep 9, 12:53 pm, Aaron <[EMAIL PROTECTED]> wrote:
> ok my problem is that I have the user upload a image file this image
> file on the page is a picture of them.
>
> What I want to do is when the mouse is over the image I want to fade
> in a menu this menu would contain buttons for editing the which photo
> to show and many other things about the image.
>
> the problem I run into is I seek a if statement. I want the mouse over
> the image to when going off meaning mouseout I want to fade out the
> menu only if the mouse is not on the menu.
>
> If I just put a mouseover and a mouseout on this image it would fade
> in the menu and when you try to put the mouse on the menu it would
> fade out.
>
> SO I just made the mouseover and mouseout for both the menu and the
> persons image.
> This caused a bad effect. When you put the mouse over the persons
> image it would fade in the menu and when you go off the image to go on
> the menu the menu would fade out and then fade back in.
>
> I want it in a way where when my mouse leaves the persons image that
> if the mouse is on the menu it would fade out at all and only fade out
> only if the mouse is not on the persons image or the menu.
>
> How can I do this ??? can you give an example???
