The dialog takes your original markup and turns it into a dialog. When
you change the dialog markup with ajax content, you're actually
overwriting the original markup. To avoid this, you could place a <div
class="feedback"> or something similar inside your dialog markup to
contain you ajax feedback, and hide/show the form / feedback depending
on the status of the process.

...Just an idea :)

On Mon, Feb 8, 2010 at 11:00 AM, West415 <malik.robin...@gmail.com> wrote:
>
> Hi,
>
> I am using a a jquery dialog box and it works fine for the most part except
> I'm running into one problem.  So here is the flow.
>
> 1. User clicks a hyperlink and dialog is opened with a form
> 2. User submits the form in the dialog via ajax and in theajax callback the
> contents of the form are replaced with a thank you/confirmation message.
> 3. User clicks close on the dialog box
>
> Now, next time I click that link, it displays the thank you/confirmation box
> and not the form.   Whenever the link is clicked it should always display
> the form, but it's like it's caching it and not calling function that loads
> the form.
>
> Any idea?
>
> West415
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-- 
Best regards,
Balázs Suhajda

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