How do I fade out all the images inside the class bMenu that are not #b2 with jQuery? Thanks.
<div class="bMenu" id="b1"><img src='b1.jpg'></div>
<div class="bMenu" id="b2"><img src='b2.jpg'></div>
<div class="bMenu" id="b3"><img src='b3.jpg'></div>
views:
65answers:
5the DIV means it's the CONTEXT of IMG (second parameter to jQuery function), so this answer should work also as yours gnarf :)
Juraj Blahunka
2010-01-21 22:03:41
I edited my post without noting it, sorry about that. So gnarf's comment was correct..
Per Holmäng
2010-01-21 22:15:48
@juraj - yeah he had `$('img:not(#b2)', '.bMenu');` when I posted that comment :)
gnarf
2010-01-21 22:21:20
@grarf :-D that's why Edit changes should be sometimes appended without allowing rewrite
Juraj Blahunka
2010-01-21 22:30:19
+2
A:
Literal answer:
$(".bMenuL:not(#b2) img").fadeOut();
Assuming you want to make sure that the #b2 img is shown as well:
$("#b2 img").fadeIn();
gnarf
2010-01-21 21:06:34
A:
Try this:
#('.bMenu > img').each(function(it){
if(it.attr('id') != 'b2'){
it.fadeOut();
}
});
Might be a way to do it with pure selectors but this should work.
Added Later:
Ok, I went and did a test... here is what I came up with:
$('div[id!=b2].bMenu > img').each(function(){
$(this).fadeOut();
});
This selector will return two images, not the one with b2.
cjstehno
2010-01-21 21:06:58
+1
A:
Get it all done at once with chaining:
$("#b2 img").show().parent().siblings(".bMenu").find("img").fadeOut();
Jonathan Sampson
2010-01-21 21:50:26