Using jQuery, how do you find only visible elements and leave hidden elements alone?

问题

So I start with items 1-4:

<div class="someDiv bold italic" style="display: none;">Lorem</div>
<div class="someDiv regular italic" style="display: block;">Lorem</div>
<div class="someDiv bold" style="display: none;">Ipsum</div>
<div class="someDiv regular" style="display: block;">Ipsum</div>

Then I have some input checkboxes:

<input class="regular" type="checkbox" />
<input class="bold" type="checkbox" />
<input class="italic" type="checkbox" />

So basically I have jQuery showing and hiding divs. Now I have another function that must iterate through these divs (one for each checkbox), and show/hide based on another criteria. But I don't want the already hidden divs to be shown again.

$(".someDiv").each(function(){
  if($(this).hasClass("regular")){
    $(this).show();
  } else {
    $(this).hide();
  };

In this example, the only remaining div should be the last div. Unfortunately, this code will make the second and fourth divs shown.

This code is very basic example of all the filters I'm going to be applying, adding etc.

回答1:

You can use the :visible selector to find only visible.

$(".someDiv:visible").each(....);

You can use the .not() selector to find only hidden.

$(".someDiv").not(":visible").each(....);

I think you can perform the same operation in your code with this one line.

$(".someDiv").hide().find(".regular").show();

Find all .someDiv and hide them, then find those with a .regular class and show them.

回答2:

You could use :visible selector to select the .someDiv that are visible.

$(".someDiv:visible").each(function(){
 if($(this).hasClass("regular")){
    $(this).show();
  } else {
    $(this).hide();
  }
});

Here is another funny way utilizing the chaining :) and making it single line.

$('.someDiv:visible').not($('.someDiv.regular:visible')).hide();

回答3:

You could do this two ways: You could add another class for the display: none elements and make them invisible via css, or you could find out the css property via jquery

via css class

html

<div class="someDiv bold italic hidden" >Lorem</div>
<div class="someDiv regular italic" >Lorem</div>
<div class="someDiv bold hidden" >Ipsum</div>
<div class="someDiv regular" >Ipsum</div>

css

.someDiv{
    display: block;
}

.hidden{
    display: none;
}

js

$(".someDiv").each(function(){
  if($(this).hasClass("hidden")){
    $(this).show();
  } else {
    $(this).hide();
  };

via jquery

$(".someDiv:visible").each(function(){
 if($(this).hasClass("regular")){
    $(this).show();
  } else {
    $(this).hide();
  }
});

回答4:

You can use the :not() selector for that and filter the results before going into the .each():

$(".someDiv:not(:hidden)").each(function(){

来源：https://stackoverflow.com/questions/16782925/using-jquery-how-do-you-find-only-visible-elements-and-leave-hidden-elements-al