[jQuery] Re: .fadeOut In mutiple elements don't work. why? what's wrong?

2007-07-06 Thread Jean
Jquery is very good for this, will adapt u into the rules \o/ On 7/4/07, Alexsandro_xpt [EMAIL PROTECTED] wrote: Exactly, I getting into standards(WC3) now. Uhm... now I understand. Thank you a lot for help me... :) bye. On 4 jul, 14:02, Ganeshji Marwaha [EMAIL PROTECTED] wrote: Glad

[jQuery] Re: .fadeOut In mutiple elements don't work. why? what's wrong?

2007-07-04 Thread Dave Probert
Sorry mate, but you are breaking the rules of HTML (as laid down by higher powers than us), as was mentioned above - ID's are individual items on a page - There should NEVER be more than one with the same ID. These are Unique ID's. You will need to go back to your original design and change it

[jQuery] Re: .fadeOut In mutiple elements don't work. why? what's wrong?

2007-07-04 Thread Eric Crull
Well, then Ganeshji is correct. Every id on a page has to be unique. In your example you have 3 ids, all named btcontrole. You can only have one named btcontrole. Change the ids of the other two. The browser is ignoring the ids of the anything after the first span which has already registered

[jQuery] Re: .fadeOut In mutiple elements don't work. why? what's wrong?

2007-07-04 Thread Alexsandro_xpt
Uhmm, now I understand. But, It's stranger, I have to use same named class design in two spans and one table, I can't make it, cos that two span and table is diferent layout. The table contein some HTML button panel, and span has two label colls and those must be fadeOut when new form fadeIn

[jQuery] Re: .fadeOut In mutiple elements don't work. why? what's wrong?

2007-07-04 Thread Ganeshji Marwaha
Glad that you are getting into standards now. Again, i understand ur concern. You are saying that classes should be used only for styling... But it is not true. Truly speaking classes are a hook left in the HTML for other languages to take advantage. One such language is CSS, and javascript can

[jQuery] Re: .fadeOut In mutiple elements don't work. why? what's wrong?

2007-07-03 Thread Ganeshji Marwaha
this is because, according to HTML spec, more than 1 element cant have the same id attribute. So, when u use $(#some) selector in jquery, it returns only the first one in the document. So, try using a class attribute instead of id. That should solve ur problem. -GTG On 7/3/07, Alexsandro_xpt

[jQuery] Re: .fadeOut In mutiple elements don't work. why? what's wrong?

2007-07-03 Thread Eric Crull
so, for example, if your code is: div id=btcontrole span id=span1hi/span span id=span2bye/span Table ...table info... /table /div Your jquery object $(#btcontrole) only contains one element, so the each() only works on it. If you wanted to have a jquery object that consists of the spans you

[jQuery] Re: .fadeOut In mutiple elements don't work. why? what's wrong?

2007-07-03 Thread Alexsandro_xpt
=/ Sorry, but I can't have objects contains like that.. My HTML is like that: span id=btcontrolehi/span span id=btcontrolebye/span . . . . . . . Table id=btcontrole ...table info... /table thank you.