If your form is in the div you update, just send back the form with the modification. Your original form will be updated by the one you send back..
This is exactly what append when you submit a form that has error in it, you send back the form with the errors tags ..and this form replaces the one you send. All this, as long as you wrapup your form in the div you update .. hope it's "clearer" .. On May 16, 3:17 pm, Dovdimus Prime <[EMAIL PROTECTED]> wrote: > This is the bit I don't get. > > The ajax->form is already coded in the view, with the $options array > all set up. > > Then the user clicks a submit button, an ajax request is made, the > logic takes place on the server. The logic now needs to somehow > conditionally modify the setup of the ajax->form, which obviously > already exists since the user is using it to make the ajax request... > > So the question is, how do I dynamically MODIFY the javascript run by > the ajax->form, as specified in the $options array? > > Does my question make sense? > > David --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "CakePHP" 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 -~----------~----~----~----~------~----~------~--~---
