[jQuery] Re: how to the pass element to a plugin ?

2007-05-31 Thread Rob Desbois 

Olivier,

The .siblings() function excludes the elements selected, e.g.:
  $(#myId).siblings();
will select all siblings of the element with ID myId, and will exclude
that element from the results.

--rob

On 5/30/07, Olivier Percebois-Garve [EMAIL PROTECTED] wrote:


 Thanks a lot.I'll test that tomorrow back at work.It looks correct.
I never used children(), and its seems to solve issues I had on other
scripts.

I also dream on a brothers() method (all other siblings except itself) ...

Olivier

Rob Desbois wrote:

Yes you're absolutely correct.

Try this:

 jQuery.fn.accordionQuizz = function(accordion) {
   this.children('[EMAIL PROTECTED], label').each(function(){


I haven't tested it though.
--rob


On 5/30/07, Olivier Percebois-Garve  [EMAIL PROTECTED] wrote:

 I dont get it working properly. I tried:

   $('#accordion').accordionQuizz();

 jQuery.fn.accordionQuizz = function(accordion){
   jQuery(this+' [EMAIL PROTECTED], '+this+'
 label').each(function(){



 I guess that here this is returning the object,  wheras I need the
 selector name.
 the plugin should apply to to radio inputs and their labels within an
 element such as #accordion.

 Olivier


 On 5/30/07, Rob Desbois  [EMAIL PROTECTED] wrote:
 
  Olivier,
 
  If you call $(...).accordionQuizz() then the jQuery object (result of
  $(...)) is accessible via the 'this' object in your function:
 
  jQuery.fn.accordionQuizz = function(accordion) {
  
  alert(this.length); // use the jQuery object
  
 
  If you call that with:
 
   $('#accordion').accordionQuizz();
 
  then in your accordionQuizz() function, 'this' is the same as
  $(#accordion)
 
   --rob
 
 
  On 5/30/07, Olivier Percebois-Garve  [EMAIL PROTECTED] wrote:
 
   Hi
  
   I am calling my plugin this way :
  
   $().accordionQuizz('accordion');
  
   my plugin gets the param this way :
  
   jQuery.fn.accordionQuizz = function(accordion){
 jQuery(accordion+' [EMAIL PROTECTED], #'+accordion+'
   label').each(function(){
  
  
   How to do in order to call the plugin this way :
  
   $('#accordion').accordionQuizz();
  
  
  
   thanks
  
  
   Olivier
  
 
 
 
  --
  Rob Desbois
  Eml: [EMAIL PROTECTED]
  Tel: 01452 760631
  Mob: 07946 705987
  There's a whale there's a whale there's a whale fish he cried, and
  the whale was in full view.
  ...Then ooh welcome. Ahhh. Ooh mug welcome.





--
Rob Desbois
Eml: [EMAIL PROTECTED]
Tel: 01452 760631
Mob: 07946 705987
There's a whale there's a whale there's a whale fish he cried, and the
whale was in full view.
...Then ooh welcome. Ahhh. Ooh mug welcome.






--
Rob Desbois
Eml: [EMAIL PROTECTED]
Tel: 01452 760631
Mob: 07946 705987
There's a whale there's a whale there's a whale fish he cried, and the
whale was in full view.
...Then ooh welcome. Ahhh. Ooh mug welcome.

[jQuery] Re: how to the pass element to a plugin ?

2007-05-31 Thread Olivier Percebois-Garve 

Hi rob

It seems to me that children() does not accept a params separated with a
comma
would that make sense ?

On 5/31/07, Rob Desbois [EMAIL PROTECTED] wrote:


Olivier,

The .siblings() function excludes the elements selected, e.g.:
   $(#myId).siblings();
will select all siblings of the element with ID myId, and will exclude
that element from the results.

--rob

On 5/30/07, Olivier Percebois-Garve [EMAIL PROTECTED] wrote:

  Thanks a lot.I'll test that tomorrow back at work.It looks correct.
 I never used children(), and its seems to solve issues I had on other
 scripts.

 I also dream on a brothers() method (all other siblings except itself)
 ...

 Olivier

 Rob Desbois wrote:

 Yes you're absolutely correct.

 Try this:

  jQuery.fn.accordionQuizz = function(accordion) {
this.children('[EMAIL PROTECTED], label').each(function(){
 

 I haven't tested it though.
 --rob


 On 5/30/07, Olivier Percebois-Garve  [EMAIL PROTECTED] wrote:
 
  I dont get it working properly. I tried:
 
$('#accordion').accordionQuizz();
 
  jQuery.fn.accordionQuizz = function(accordion){
jQuery(this+' [EMAIL PROTECTED], '+this+'
  label').each(function(){
 
 
 
  I guess that here this is returning the object,  wheras I need the
  selector name.
  the plugin should apply to to radio inputs and their labels within an
  element such as #accordion.
 
  Olivier
 
 
  On 5/30/07, Rob Desbois  [EMAIL PROTECTED] wrote:
  
   Olivier,
  
   If you call $(...).accordionQuizz() then the jQuery object (result
   of $(...)) is accessible via the 'this' object in your function:
  
   jQuery.fn.accordionQuizz = function(accordion) {
   
   alert(this.length); // use the jQuery object
   
  
   If you call that with:
  
$('#accordion').accordionQuizz();
  
   then in your accordionQuizz() function, 'this' is the same as
   $(#accordion)
  
--rob
  
  
   On 5/30/07, Olivier Percebois-Garve  [EMAIL PROTECTED] wrote:
  
Hi
   
I am calling my plugin this way :
   
$().accordionQuizz('accordion');
   
my plugin gets the param this way :
   
jQuery.fn.accordionQuizz = function(accordion){
  jQuery(accordion+' [EMAIL PROTECTED], #'+accordion+'
label').each(function(){
   
   
How to do in order to call the plugin this way :
   
$('#accordion').accordionQuizz();
   
   
   
thanks
   
   
Olivier
   
  
  
  
   --
   Rob Desbois
   Eml:  [EMAIL PROTECTED]
   Tel: 01452 760631
   Mob: 07946 705987
   There's a whale there's a whale there's a whale fish he cried, and
   the whale was in full view.
   ...Then ooh welcome. Ahhh. Ooh mug welcome.
 
 
 


 --
 Rob Desbois
 Eml: [EMAIL PROTECTED]
 Tel: 01452 760631
 Mob: 07946 705987
 There's a whale there's a whale there's a whale fish he cried, and the
 whale was in full view.
 ...Then ooh welcome. Ahhh. Ooh mug welcome.





--
Rob Desbois
Eml: [EMAIL PROTECTED]
Tel: 01452 760631
Mob: 07946 705987
There's a whale there's a whale there's a whale fish he cried, and the
whale was in full view.
...Then ooh welcome. Ahhh. Ooh mug welcome.

[jQuery] Re: how to the pass element to a plugin ?

2007-05-30 Thread Rob Desbois 

Olivier,

If you call $(...).accordionQuizz() then the jQuery object (result of
$(...)) is accessible via the 'this' object in your function:

jQuery.fn.accordionQuizz = function(accordion) {



  alert(this.length); // use the jQuery object




If you call that with:


$('#accordion').accordionQuizz();


then in your accordionQuizz() function, 'this' is the same as
$(#accordion)

--rob


On 5/30/07, Olivier Percebois-Garve [EMAIL PROTECTED] wrote:


Hi

I am calling my plugin this way :

$().accordionQuizz('accordion');

my plugin gets the param this way :

jQuery.fn.accordionQuizz = function(accordion){
  jQuery(accordion+' [EMAIL PROTECTED], #'+accordion+'
label').each(function(){


How to do in order to call the plugin this way :

$('#accordion').accordionQuizz();



thanks


Olivier





--
Rob Desbois
Eml: [EMAIL PROTECTED]
Tel: 01452 760631
Mob: 07946 705987
There's a whale there's a whale there's a whale fish he cried, and the
whale was in full view.
...Then ooh welcome. Ahhh. Ooh mug welcome.

[jQuery] Re: how to the pass element to a plugin ?

2007-05-30 Thread Olivier Percebois-Garve 

I dont get it working properly. I tried:

 $('#accordion').accordionQuizz();

jQuery.fn.accordionQuizz = function(accordion){
 jQuery(this+' [EMAIL PROTECTED], '+this+'
label').each(function(){



I guess that here this is returning the object,  wheras I need the
selector name.
the plugin should apply to to radio inputs and their labels within an
element such as #accordion.

Olivier


On 5/30/07, Rob Desbois [EMAIL PROTECTED] wrote:


Olivier,

If you call $(...).accordionQuizz() then the jQuery object (result of
$(...)) is accessible via the 'this' object in your function:

jQuery.fn.accordionQuizz = function(accordion) {

   alert(this.length); // use the jQuery object


If you call that with:

 $('#accordion').accordionQuizz();

then in your accordionQuizz() function, 'this' is the same as
$(#accordion)

--rob


On 5/30/07, Olivier Percebois-Garve [EMAIL PROTECTED] wrote:

 Hi

 I am calling my plugin this way :

 $().accordionQuizz('accordion');

 my plugin gets the param this way :

 jQuery.fn.accordionQuizz = function(accordion){
   jQuery(accordion+' [EMAIL PROTECTED], #'+accordion+'
 label').each(function(){


 How to do in order to call the plugin this way :

 $('#accordion').accordionQuizz();



 thanks


 Olivier




--
Rob Desbois
Eml: [EMAIL PROTECTED]
Tel: 01452 760631
Mob: 07946 705987
There's a whale there's a whale there's a whale fish he cried, and the
whale was in full view.
...Then ooh welcome. Ahhh. Ooh mug welcome.

[jQuery] Re: how to the pass element to a plugin ?

2007-05-30 Thread Rob Desbois 

Yes you're absolutely correct.

Try this:


jQuery.fn.accordionQuizz = function(accordion) {
  this.children('[EMAIL PROTECTED], label').each(function(){



I haven't tested it though.
--rob


On 5/30/07, Olivier Percebois-Garve [EMAIL PROTECTED] wrote:


I dont get it working properly. I tried:

  $('#accordion').accordionQuizz();

jQuery.fn.accordionQuizz = function(accordion){
  jQuery(this+' [EMAIL PROTECTED], '+this+'
label').each(function(){



I guess that here this is returning the object,  wheras I need the
selector name.
the plugin should apply to to radio inputs and their labels within an
element such as #accordion.

Olivier


On 5/30/07, Rob Desbois [EMAIL PROTECTED] wrote:

 Olivier,

 If you call $(...).accordionQuizz() then the jQuery object (result of
 $(...)) is accessible via the 'this' object in your function:

 jQuery.fn.accordionQuizz = function(accordion) {
 
alert(this.length); // use the jQuery object
 

 If you call that with:

  $('#accordion').accordionQuizz();

 then in your accordionQuizz() function, 'this' is the same as
 $(#accordion)

 --rob


 On 5/30/07, Olivier Percebois-Garve [EMAIL PROTECTED] wrote:

  Hi
 
  I am calling my plugin this way :
 
  $().accordionQuizz('accordion');
 
  my plugin gets the param this way :
 
  jQuery.fn.accordionQuizz = function(accordion){
jQuery(accordion+' [EMAIL PROTECTED], #'+accordion+'
  label').each(function(){
 
 
  How to do in order to call the plugin this way :
 
  $('#accordion').accordionQuizz();
 
 
 
  thanks
 
 
  Olivier
 



 --
 Rob Desbois
 Eml: [EMAIL PROTECTED]
 Tel: 01452 760631
 Mob: 07946 705987
 There's a whale there's a whale there's a whale fish he cried, and the
 whale was in full view.
 ...Then ooh welcome. Ahhh. Ooh mug welcome.






--
Rob Desbois
Eml: [EMAIL PROTECTED]
Tel: 01452 760631
Mob: 07946 705987
There's a whale there's a whale there's a whale fish he cried, and the
whale was in full view.
...Then ooh welcome. Ahhh. Ooh mug welcome.

[jQuery] Re: how to the pass element to a plugin ?

2007-05-30 Thread Olivier Percebois-Garve 

Thanks a lot.I'll test that tomorrow back at work.It looks correct.
I never used children(), and its seems to solve issues I had on other 
scripts.


I also dream on a brothers() method (all other siblings except itself) ...

Olivier

Rob Desbois wrote:

Yes you're absolutely correct.

Try this:

jQuery.fn.accordionQuizz = function(accordion) {
  this.children('[EMAIL PROTECTED], label').each(function(){



I haven't tested it though.
--rob


On 5/30/07, *Olivier Percebois-Garve*  [EMAIL PROTECTED] 
mailto:[EMAIL PROTECTED] wrote:


I dont get it working properly. I tried:

  $('#accordion').accordionQuizz();

jQuery.fn.accordionQuizz = function(accordion){
  jQuery(this+' [EMAIL PROTECTED], '+this+'
label').each(function(){




I guess that here this is returning the object,  wheras I need
the selector name.
the plugin should apply to to radio inputs and their labels within
an element such as #accordion.

Olivier



On 5/30/07, *Rob Desbois*  [send email to [EMAIL PROTECTED]
via gmail] [EMAIL PROTECTED] mailto:[EMAIL PROTECTED]
wrote:

Olivier,

If you call $(...).accordionQuizz() then the jQuery object
(result of $(...)) is accessible via the 'this' object in your
function:

jQuery.fn.accordionQuizz = function(accordion) {

   alert(this.length); // use the jQuery object


If you call that with:

$('#accordion').accordionQuizz();

then in your accordionQuizz() function, 'this' is the same as
$(#accordion)

--rob



On 5/30/07, *Olivier Percebois-Garve*  [send email to
[EMAIL PROTECTED] via gmail] [EMAIL PROTECTED]
mailto:[EMAIL PROTECTED] wrote:

Hi

I am calling my plugin this way :

$().accordionQuizz('accordion');

my plugin gets the param this way :

jQuery.fn.accordionQuizz = function(accordion){
  jQuery(accordion+' [EMAIL PROTECTED],

#'+accordion+' label').each(function(){


How to do in order to call the plugin this way :

$('#accordion').accordionQuizz();



thanks


Olivier




-- 
Rob Desbois

Eml: [send email to [EMAIL PROTECTED] via gmail]
[EMAIL PROTECTED] mailto:[EMAIL PROTECTED]
Tel: 01452 760631
Mob: 07946 705987
There's a whale there's a whale there's a whale fish he
cried, and the whale was in full view.
...Then ooh welcome. Ahhh. Ooh mug welcome. 






--
Rob Desbois
Eml: [EMAIL PROTECTED] mailto:[EMAIL PROTECTED]
Tel: 01452 760631
Mob: 07946 705987
There's a whale there's a whale there's a whale fish he cried, and 
the whale was in full view.
...Then ooh welcome. Ahhh. Ooh mug welcome.