subject:"\[jQuery\] Re\: question about changes to \:not\(\) in 1.3"

[jQuery] Re: question about changes to :not() in 1.3

2009-01-19 Thread jdwbell


Thanks for the reply.  That would give me every non-div element which
contained a div.  In the original example I was looking for every div
which did not contain another div plus every other non-div element.
Based on John's reply I'm guessing there is bug since $(':not(div:has
(div))') is supposed to be equivalent to
$('*').not('div:has(div)') but doesn't seem to be in the example I've
given.


On Jan 17, 2:20 pm, Ricardo Tomasi ricardob...@gmail.com wrote:
 Shouldn't it simply be

 $(':not(div):has(div)') ?

 Note that it gives you all the chain of parents to the element which
 contains a DIV.

 On Jan 16, 6:26 pm, jdwbell jdwb...@gmail.com wrote:

  Thank you very much for your reply.  I am finding that with the
  example listed below these two work as expected (that is they are
  returning non divs and all divs which donotcontain other divs):

  $('*').not('div:has(div)')
  $('*').not($('div:has(div)'))

  but these three:

  $(':not(div:has(div))')
  $('*:not(div:has(div))'
  $('*').filter(':not(div:has(div))')

  are excluding ALL divs as all other elements which contain divs

  Hopefully that explanation makes because I'm kind of confusing
  myself!  I've tried this on IE 6 and Chrome with the same results.

  Thanks!

  html
    head
      style
        div, fieldset, p {
          border: ridge 1px silver;
          padding: 20px;
          margin: 20px;
        }
        .wrappedElement {
          border: ridge 2px #FF;
        }
       /style
       script type=text/javascript
          src=../../scripts/jquery-1.3.js/script
       script
         $(function(){
           $(':not(div:has(div))').addClass('wrappedElement');
           //$('*:not(div:has(div))').addClass('wrappedElement');
           //$('*').filter(':not(div:has(div))').addClass('wrappedElement');
           //$('*').not('div:has(div)').addClass('wrappedElement');
           //$('*').not($('div:has(div)')).addClass('wrappedElement');
         })
       /script
    /head
    body
      div
        div
          div
            pparagraph surrounded by three divs/p
          /div
        /div
       /div
       fieldset
         div
           div surrounded by fieldset
         /div
       /fieldset
     /body
  /html

  On Jan 16, 1:50 pm, John Resig jere...@gmail.com wrote:

   $(':not(div:has(div))') is equivalent to
   $('*:not(div:has(div))') is equivalent to
   $('*').filter(':not(div:has(div))') is equivalent to
   $('*').not('div:has(div)')

   Hope that helps to answer your question :)

   --John

   On Fri, Jan 16, 2009 at 11:24 AM, jdwbell jdwb...@gmail.com wrote:

Here I am trying to get every element which isnota div that contains
another div.

Is $(':not(div:has(div))') supposed to now be equivalent to $('*').not
($('div:has(div)')) or am I misunderstanding what the change was (I'm
just getting started with jQuery)?

Thanks!

[jQuery] Re: question about changes to :not() in 1.3

2009-01-19 Thread Ricardo Tomasi


sorry for the misinterpretation. Yeah, that seems like a bug.

$(':not(div:has(div))') gave the exact same results as $('*').not
('div:has(div)') in jQuery 1.2.6.

In 1.3 $(':not(div:has(div))') excludes html, body and any div
elements.

On Jan 19, 7:07 pm, jdwbell jdwb...@gmail.com wrote:
 Thanks for the reply.  That would give me every non-div element which
 contained a div.  In the original example I was looking for every div
 which did not contain another div plus every other non-div element.
 Based on John's reply I'm guessing there is bug since $(':not(div:has
 (div))') is supposed to be equivalent to
 $('*').not('div:has(div)') but doesn't seem to be in the example I've
 given.

 On Jan 17, 2:20 pm, Ricardo Tomasi ricardob...@gmail.com wrote:

  Shouldn't it simply be

  $(':not(div):has(div)') ?

  Note that it gives you all the chain of parents to the element which
  contains a DIV.

  On Jan 16, 6:26 pm, jdwbell jdwb...@gmail.com wrote:

   Thank you very much for your reply.  I am finding that with the
   example listed below these two work as expected (that is they are
   returning non divs and all divs which donotcontain other divs):

   $('*').not('div:has(div)')
   $('*').not($('div:has(div)'))

   but these three:

   $(':not(div:has(div))')
   $('*:not(div:has(div))'
   $('*').filter(':not(div:has(div))')

   are excluding ALL divs as all other elements which contain divs

   Hopefully that explanation makes because I'm kind of confusing
   myself!  I've tried this on IE 6 and Chrome with the same results.

   Thanks!

   html
     head
       style
         div, fieldset, p {
           border: ridge 1px silver;
           padding: 20px;
           margin: 20px;
         }
         .wrappedElement {
           border: ridge 2px #FF;
         }
        /style
        script type=text/javascript
           src=../../scripts/jquery-1.3.js/script
        script
          $(function(){
            $(':not(div:has(div))').addClass('wrappedElement');
            //$('*:not(div:has(div))').addClass('wrappedElement');
            //$('*').filter(':not(div:has(div))').addClass('wrappedElement');
            //$('*').not('div:has(div)').addClass('wrappedElement');
            //$('*').not($('div:has(div)')).addClass('wrappedElement');
          })
        /script
     /head
     body
       div
         div
           div
             pparagraph surrounded by three divs/p
           /div
         /div
        /div
        fieldset
          div
            div surrounded by fieldset
          /div
        /fieldset
      /body
   /html

   On Jan 16, 1:50 pm, John Resig jere...@gmail.com wrote:

$(':not(div:has(div))') is equivalent to
$('*:not(div:has(div))') is equivalent to
$('*').filter(':not(div:has(div))') is equivalent to
$('*').not('div:has(div)')

Hope that helps to answer your question :)

--John

On Fri, Jan 16, 2009 at 11:24 AM, jdwbell jdwb...@gmail.com wrote:

 Here I am trying to get every element which isnota div that contains
 another div.

 Is $(':not(div:has(div))') supposed to now be equivalent to $('*').not
 ($('div:has(div)')) or am I misunderstanding what the change was (I'm
 just getting started with jQuery)?

 Thanks!

[jQuery] Re: question about changes to :not() in 1.3

2009-01-17 Thread Ricardo Tomasi


Shouldn't it simply be

$(':not(div):has(div)') ?

Note that it gives you all the chain of parents to the element which
contains a DIV.

On Jan 16, 6:26 pm, jdwbell jdwb...@gmail.com wrote:
 Thank you very much for your reply.  I am finding that with the
 example listed below these two work as expected (that is they are
 returning non divs and all divs which do not contain other divs):

 $('*').not('div:has(div)')
 $('*').not($('div:has(div)'))

 but these three:

 $(':not(div:has(div))')
 $('*:not(div:has(div))'
 $('*').filter(':not(div:has(div))')

 are excluding ALL divs as all other elements which contain divs

 Hopefully that explanation makes because I'm kind of confusing
 myself!  I've tried this on IE 6 and Chrome with the same results.

 Thanks!

 html
   head
     style
       div, fieldset, p {
         border: ridge 1px silver;
         padding: 20px;
         margin: 20px;
       }
       .wrappedElement {
         border: ridge 2px #FF;
       }
      /style
      script type=text/javascript
         src=../../scripts/jquery-1.3.js/script
      script
        $(function(){
          $(':not(div:has(div))').addClass('wrappedElement');
          //$('*:not(div:has(div))').addClass('wrappedElement');
          //$('*').filter(':not(div:has(div))').addClass('wrappedElement');
          //$('*').not('div:has(div)').addClass('wrappedElement');
          //$('*').not($('div:has(div)')).addClass('wrappedElement');
        })
      /script
   /head
   body
     div
       div
         div
           pparagraph surrounded by three divs/p
         /div
       /div
      /div
      fieldset
        div
          div surrounded by fieldset
        /div
      /fieldset
    /body
 /html

 On Jan 16, 1:50 pm, John Resig jere...@gmail.com wrote:

  $(':not(div:has(div))') is equivalent to
  $('*:not(div:has(div))') is equivalent to
  $('*').filter(':not(div:has(div))') is equivalent to
  $('*').not('div:has(div)')

  Hope that helps to answer your question :)

  --John

  On Fri, Jan 16, 2009 at 11:24 AM, jdwbell jdwb...@gmail.com wrote:

   Here I am trying to get every element which is not a div that contains
   another div.

   Is $(':not(div:has(div))') supposed to now be equivalent to $('*').not
   ($('div:has(div)')) or am I misunderstanding what the change was (I'm
   just getting started with jQuery)?

   Thanks!

[jQuery] Re: question about changes to :not() in 1.3

2009-01-16 Thread John Resig


$(':not(div:has(div))') is equivalent to
$('*:not(div:has(div))') is equivalent to
$('*').filter(':not(div:has(div))') is equivalent to
$('*').not('div:has(div)')

Hope that helps to answer your question :)

--John



On Fri, Jan 16, 2009 at 11:24 AM, jdwbell jdwb...@gmail.com wrote:

 Here I am trying to get every element which is not a div that contains
 another div.

 Is $(':not(div:has(div))') supposed to now be equivalent to $('*').not
 ($('div:has(div)')) or am I misunderstanding what the change was (I'm
 just getting started with jQuery)?

 Thanks!

[jQuery] Re: question about changes to :not() in 1.3

2009-01-16 Thread jdwbell


Thank you very much for your reply.  I am finding that with the
example listed below these two work as expected (that is they are
returning non divs and all divs which do not contain other divs):

$('*').not('div:has(div)')
$('*').not($('div:has(div)'))

but these three:

$(':not(div:has(div))')
$('*:not(div:has(div))'
$('*').filter(':not(div:has(div))')

are excluding ALL divs as all other elements which contain divs

Hopefully that explanation makes because I'm kind of confusing
myself!  I've tried this on IE 6 and Chrome with the same results.

Thanks!


html
  head
style
  div, fieldset, p {
border: ridge 1px silver;
padding: 20px;
margin: 20px;
  }
  .wrappedElement {
border: ridge 2px #FF;
  }
 /style
 script type=text/javascript
src=../../scripts/jquery-1.3.js/script
 script
   $(function(){
 $(':not(div:has(div))').addClass('wrappedElement');
 //$('*:not(div:has(div))').addClass('wrappedElement');
 //$('*').filter(':not(div:has(div))').addClass('wrappedElement');
 //$('*').not('div:has(div)').addClass('wrappedElement');
 //$('*').not($('div:has(div)')).addClass('wrappedElement');
   })
 /script
  /head
  body
div
  div
div
  pparagraph surrounded by three divs/p
/div
  /div
 /div
 fieldset
   div
 div surrounded by fieldset
   /div
 /fieldset
   /body
/html

On Jan 16, 1:50 pm, John Resig jere...@gmail.com wrote:
 $(':not(div:has(div))') is equivalent to
 $('*:not(div:has(div))') is equivalent to
 $('*').filter(':not(div:has(div))') is equivalent to
 $('*').not('div:has(div)')

 Hope that helps to answer your question :)

 --John

 On Fri, Jan 16, 2009 at 11:24 AM, jdwbell jdwb...@gmail.com wrote:

  Here I am trying to get every element which is not a div that contains
  another div.

  Is $(':not(div:has(div))') supposed to now be equivalent to $('*').not
  ($('div:has(div)')) or am I misunderstanding what the change was (I'm
  just getting started with jQuery)?

  Thanks!

[jQuery] Re: question about changes to :not() in 1.3

[jQuery] Re: question about changes to :not() in 1.3

[jQuery] Re: question about changes to :not() in 1.3

[jQuery] Re: question about changes to :not() in 1.3

[jQuery] Re: question about changes to :not() in 1.3

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