jQuery:给定一个选择器,只找到它的可见元素
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jQuery: Given a selector, find only its visible elements
提问by MegaMatt
This should be an easy one. I have a variable that I've already declared called $listItems. The declaration looks like this:
这应该是一件容易的事。我有一个已经声明为 $listItems 的变量。声明如下所示:
var $listItems = $ul.children('li'); // $ul is just a selected unordered list
Later in my code, I'd like to only get the ones that are currently visible. How would I go about that? Something like:
稍后在我的代码中,我只想获取当前可见的那些。我该怎么做?就像是:
$listItems.parent().children(':visible')?
Thanks.
谢谢。
回答by Nick Craver
回答by Jason McCreary
You have it with the :visibleselector. It can be used in any of the jQuery collection methods $()
, filter()
, children()
, find()
, etc.
您可以使用:visible选择器。它可以在任何的jQuery的收集方法使用$()
,filter()
,children()
,find()
,等。
Note:There is a difference between something that is visibleon the page and has its visibility
property set.
注意:页面上可见的内容与visibility
设置其属性的内容之间存在差异。