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

Question

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.

Answer 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.

Answer 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();