I have a HTML structure that contains nested divs with开发者_高级运维 the container
class:
<div class="container" id="0">
<div id="1">
<div class="container" id="2">
<div class="container" id="3"></div>
</div>
<div id="4">
<div class="container" id="5"></div>
</div>
</div>
</div>
This could include more divs and deeper/different nesting.
Starting from some point in the tree I'd like to find all containers in that subtree that are not nested within other containers. So for example from div #1
I'd like to find the divs #2
and #5
, but not #3
(since it is nested in container #2
already found).
What would be the best way to accomplish this?
function getOuterContainers(el) {
if (el.length == 0)
return $([]);
else
return el.children('.container').add( getOuterContainers(el.children(':not(.container)')) );
}
check it out: http://jsfiddle.net/GZ3Tx/
Try using jQuery's filter()
method:
$items = $('#0').find('.container').filter(function(index) {
return !$(this).parents(':not(#0)').hasClass('container');
});
Basically, it finds all items with class .container
below your top .container
and filters the set based on whether or not the direct parent has the class .container
.
See here for a working example: http://jsfiddle.net/TzL8Z/1
A solution working with arbitrary starting nodes:
function getContainers(base) {
return base.find('.container').not(base.find('.container .container'));
}
Example: http://jsfiddle.net/tejp/GSRH2/2/
The advantage is that this works with an arbitrary starting node that doesn't need to be selectable with a simple JQuery selector, like it is the case in some of the other answers. Also it doesn't require "manually" walking over the whole tree with children()
.
The approach can also be easily extended to similar problems, like for example getting all .item
s not nested into a .container
.
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