如何删除jQuery组的其他成员的后代元素？

Question

Suppose I've collected a jQuery group of DOM nodes matching some selector, and the selector is such that it's possible that some of the matching nodes might be descendants of other matches, eg:

<div class="row">
    <!-- ... -->
    <div class="row"><!-- ... --></div>
    <!-- ... -->
</div>
<div class="row">
    <!-- ... -->
    <div class="row">
        <!-- ... -->
        <div class="row"><!-- ... --></div>
    </div>
    <div class="row"><!-- ... --></div>
    <!-- ... -->
</div>

$rows = $('.row');

Now suppose that the action I'm going to take will affect each node's entire subtree, and I need to be sure that I act on any given DOM node at most once. How would I remove all elements of my group that are descendants of other members of the same group?

Answer 1

The simplest and cleanest way I've found to accomplish this is to combine the behaviors of $.fn.not() and $.fn.find() :

var $rows = $('.row');
$rows = $rows.not($rows.find($rows));

$elems.find($elems) will return all members of $elems that are descendants of one or more members of $elems (not including the element itself, so no worries there). $elems.not(group) will return a only the members of $elems that are not in group . Together, this will find all members of $elems that are children of other members, then return everything else.