jQuery detect visible but hidden elements

问题

This seems like it should be fairly easy - but I can't find the right selector for it

According to the docs (http://api.jquery.com/hidden-selector/ and http://api.jquery.com/visible-selector/)...

Elements can be considered hidden for several reasons:

An ancestor element is hidden, so the element is not shown on the page.

What I want to detect is "this element is visible, but is contained in a hidden parent". Ie, if I made the parent visible, this element would also be visible.

回答1:

If this is something you'll commonly use, make your own selector :) Here's an example:

jQuery.expr[':'].hiddenByParent = function(a) { 
   return jQuery(a).is(':hidden') && jQuery(a).css('display') != 'none'; 
};

You can use it like this, test markup:

<div style="display: none" id="parent">
  <div>
      <div id="child">Test</div>
  </div>
</div>
​

Examples of use:

$("div:hiddenByParent").length;​​​​​​​​​​​​​​​​​​ // "2" (plain div + child match)
$("#child").is(":hiddenByParent"); // true

Alternatively, you can use the .filter() function like this:

$('selector').filter(function() {
  return $(this).is(':hidden') && $(this).css('display') != 'none';
}

回答2:

jQuery has this all built-in nowdays

$("#child").closest(':hidden').length == 0

回答3:

If it is a specific element that you are looking for then you could check it's display property

$('#element').css('display') != 'none';

If it wasn't a specific element then you could find the parent nodes that are hidden using :hidden then use a custom function to look for nodes of the type you want. E.g.

$('parent-selector:hidden').find('node-selector').each(function(){
  if($(this).css('display') != 'none') {
    // do what you wanted
  }
});

If you want a clean selector then i think that you're going to be out of luck as i don't think what you want is part of the CSS spec, so won't be there as a selector in jQuery.

来源：https://stackoverflow.com/questions/2553534/jquery-detect-visible-but-hidden-elements