Reputation:
I have a menu. When a tree is opened, it gets the class active
. I wan't to get the deepest sub-menu of the menu with the class active
.
Upvotes: 0
Views: 2445
Reputation: 16063
And a vanilla version :
function deepest (subMenu, select) {
return [].slice.call (subMenu.querySelectorAll (select)).reduce (
function (deepest, el) {
for (var d = 0, e = el; e !== subMenu; d++, e = e.parentNode);
return d > deepest.d ? {d: d, el: el} : deepest;
}, {d: 0, el: subMenu}).el;
};
call with
deepest (subMenu, '.active');
where the second parameter selects the elements to be considered.
Tested on a modified version of @Rory's fiddle
Upvotes: 1
Reputation: 337714
You can use the parents()
collection to count the depth of the element and then return the lowest for the specific selector. Try this:
<div>
1
<div class="active">1.1</div>
<div class="active">
1.2
<div class="active">
1.2.1
<div class="active">1.2.1.1</div>
</div>
<div class="active">1.2.2</div>
</div>
<div class="active">1.3</div>
<div class="active">
1.4
<div class="active">1.4.1</div>
<div class="active">1.4.2</div>
</div>
<div class="active">1.5</div>
</div>
var $elems = $('.active'), $deepest, lowestLevel = 0;
$elems.each(function() {
var depth = $(this).parents().length;
if (depth > lowestLevel) {
$deepest = $(this);
lowestLevel = depth;
}
});
$deepest.css('border', '1px solid #C00');
Upvotes: 1