Why is my hidden <tr> not really hidden?

Question

I have this simple html markup generated from classic asp:

<table>
  <tr class="trClass">
    <td>Hello </td>
  </tr>
  <tr class ="trClass">
    <td>World!</td> 
  </tr>
</table>

If i set the tr belonging to Hello to hide() using Jquery it hides. Good!

But, when i use this $('.trClass:visible').each(function() { alert('visible') }); it shows the output 'visible' twice.

Why is this?

My problem here is that im filtering a table on a column with a selection box. But after filtering i need to perform some calculations on those rows that are visible in the table, but i get all rows right now.

Any ideas?

/Daniel

Answer 1

The :visible selector does not test the display style property, you want to use :hidden instead, the 1.3.2 release notes say:

...if your element's CSS display is "none", or any of its parent/ancestor element's display is "none", or if the element's width is 0 and the element's height is 0 then an element will be reported as hidden.

These will correctly select your visible rows:

$('.trClass:not(:hidden)').each(function() { 
    alert('visible'); 
});

or:

$('.trClass').each(function() { 
    if(!$(this).is(':hidden')) {
        alert('visible'); 
    }
});

or:

$('.trClass').filter('not:(:hidden)').each(function() { 
    alert('visible');
});

hide sets the style to display="none". The release notes for jQuery 1.3.2 also say:

In jQuery 1.3.2 an element is visible if its browser-reported offsetWidth or offsetHeight is greater than 0.

so I guess in this case the :visible selector is erroneously not matching anything because the rows are not occupying any space according to the calculations performed, despite the fact that their CSS display property is not set to none. Conversely, :hidden correctly matches elements with style="display:none" so testing for non :hidden elements works just fine.