[jQuery] Re: how to the pass element to a plugin ?
Olivier,
The following works fine for me:
$(document).ready(function() {
$("#top").children(".one, .two").each(function() {
alert(this.id);
});
});
Can you post your code?
--rob
On 5/31/07, Olivier Percebois-Garve <[EMAIL PROTECTED]> wrote:
Hi rob
It seems to me that children() does not accept a params separated with a
comma
would that make sense ?
On 5/31/07, Rob Desbois < [EMAIL PROTECTED]> wrote:
>
> Olivier,
>
> The .siblings() function excludes the elements selected, e.g.:
>$("#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 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.
>
--
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 ?
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 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 ?
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 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 ?
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 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] >
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]
> 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]
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 ?
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 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 ?
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 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 ?
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.
