jQuery按类搜索且CSS值不等于foo

Question

Not sure where to look in the jQuery documentation to find this. I have a jQuery element that I found on the DOM.

Within that, I want all elements with class == 'bar' and css value 'display' == 'none' . The first part is easy:

$myElement.find(".bar");

How can I get all elements within this list where the css value 'display' == 'none' ?

All in one line if possible. Thanks!

Answer 1

You can use filter:

$myElement.find(".bar").filter(function(){
  return $(this).css('display') == 'none';
}).Apply_your_jQuery_method_now();

Answer 2

这样做：

 $myElement.find(".bar").filter(":hidden"); //you have all display:none ones

Answer 3

您可以使用:visible选择器

$myElement.find(".bar:not(:visible)");