I'd say, just write out a prototype AJAX request... its simple enough and you have full control over options.
http://www.sergiopereira.com/articles/prototype.js.html#UsingAjaxRequest http://www.sergiopereira.com/articles/prototype.js.html#UsingAjaxUpdater thanks, -alan- On May 7, 9:42 am, "[EMAIL PROTECTED]" <[EMAIL PROTECTED]> wrote: > I have a question > > I have a javascript function that in my view is inside a > $javascript->codeBlock. > > Now in this function I must use a $ajax->remoteFunction. > How can I do? > > echo $javascript->codeBlock(" > function FUNCT_A (id) { > > <!----- here I create a div called "div_$id" ----!> > > <!----- here I want use $ajax->remoteFunction to update the > content of "div_$id" ----!> > > "); --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Cake PHP" group. To post to this group, send email to [email protected] To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/cake-php?hl=en -~----------~----~----~----~------~----~------~--~---
