jQuery hover fadeIn fadeOut problem
Question
I just got back to html, javascript and jquery and i'm a bit rusty. I've mostly been doing objective-c for a while and getting back to jquery is a little difficult. I'm trying to use the jQuery fadeIn and fadeOut bit but its not working for some reason... Here's the html and js I've been working on :
<script type="text/javascript" src="http://code.jquery.com/jquery-1.6.3.min.js">
function showImg(2img) {
$(2img).fadeIn('slow', function() {
// Animation complete
});
}
function hideImg(2img) {
$(2img).fadeOut('slow', function() {
// Animation complete
});
}
</script>
<body>
<table width="1659" height="701" border="0" align="center">
<tr>
<th width="325" scope="col"><div style="position: relative; left: 0; top: 0;"><img id="img1" src="" width="232" height="232" alt="" onmouseover="showimg(2img1)" onmouseout="hideimg(2img1)" style="position: relative; top: 0; left: 0;"/>
<img id="2img1" src="" width="32" height="32" alt="" style="position: absolute; top: 100px; left: 150px; visibility:hidden;"/></div></th>
<th width="325" scope="col"><div style="position: relative; left: 0; top: 0;"><img id="img2" src="" width="232" height="232" alt="" onmouseover="showimg(2img2)" onmouseout="hideimg(2img2)" style="position: relative; top: 0; left: 0;"/>
<img id="2img2" src="" width="32" height="32" alt="" style="position: absolute; top: 100px; left: 150px; visibility:hidden;"/></div></th>
<th width="325" scope="col"><div style="position: relative; left: 0; top: 0;"><img id="img3" src="" width="232" height="232" alt="" onmouseover="showimg(2img3)" onmouseout="hideimg(2img3)" style="position: relative; top: 0; left: 0;"/>
<img id="2img3" src="" width="32" height="32" alt="" style="position: absolute; top: 100px; left: 150px; visibility:hidden;"/></div></th>
<th width="325" scope="col"><div style="position: relative; left: 0; top: 0;"><img id="img4" src="" width="232" height="232" alt="" onmouseover="showimg(2img4)" onmouseout="hideimg(2img4)" style="position: relative; top: 0; left: 0;"/>
<img id="2img4" src="" width="32" height="32" alt="" style="position: absolute; top: 100px; left: 150px; visibility:hidden;"/></div></th>
<th width="325" scope="col"><div style="position: relative; left: 0; top: 0;"><img id="img5" src="" width="232" height="232" alt="" onmouseover="showimg(2img5)" onmouseout="hideimg(2img5)" style="position: relative; top: 0; left: 0;"/>
<img id="2img5" src="" width="32" height="32" alt="" style="position: absolute; top: 100px; left: 150px; visibility:hidden;"/></div></th>
</tr>
</table>
</body>
4 answers
solution1
3 2011-09-11 16:49:22
I think you'd find it better to do this:
$(function() {
$('table > div > img:first').hover(function() {
showImg($(this).next());
}, function() {
hideImg($(this).next());
});
});
This way you wont be replicating the onmouseover and onmouseoout code in your html.
The errors in your code are - No # for the id of the images to fade, although it seems to work in chrome I don't know if it will work as expected in every browser, also the function names don't have the same case.
solution2
1 2011-09-11 16:49:03
$(2img)应该是$('#2img') ，就是这样
solution3
1 2011-09-11 16:52:52
$（2img）应为$（'＃'+ 2img）并调用showImg（“2img2”）之类的函数
solution4
0 2011-09-11 18:59:36
DEMO
$('.tb td').each(function(){
$(this).find('img').wrapAll('<div />');
$('.tb td div').css({position:'relative'});
$(this).find('img:eq(1)').css({position:'absolute', left:'0px', top:'0px'}).hide();
});
$('.tb td div').hover(function(){
$(this).find('img:eq(1)').stop().fadeTo(400,1);
},function(){
$(this).find('img:eq(1)').stop().fadeTo(400,0);
});
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