But you are just asking for a normal AJAX update, why are you trying to do
it by yourslef? DRY baby ;) Try this:
index.thtml:
?php echo $ajax-div('updateableDiv'); ?
form method=post id=finished_topics_form name=finished_topics_form
onsubmit=return false;
...
?php echo $ajax-submit('
Thanks for answer, but I wasn't asking for that. I used Ajax.Updater
before and what You have described I already used.
I will try to explain differently:
If I call my action like usual (ie. non-ajax):
http://myserver/mycontroller/myaction/
I get normal response. OK?
I want to have that
PHP
Asunto: Re: Rendering non-ajax view after calling action with ajax call
Thanks for answer, but I wasn't asking for that. I used Ajax.Updater
before and what You have described I already used.
I will try to explain differently:
If I call my action like usual (ie. non-ajax):
http://myserver
That's not what I was looking for, but thanks anyway...
Guess my language is not quite good in explaining.. :)
Anyway, I solved what I wanted...
I wanted to submit this form to a new page.
But I had some other AJAX calls in the same form that updates this form
(I am transfering some serialized
