When you render the form for second time (added the other form), and the model for the first form has been saved(), you need to put the id of the record in the form on a hidden field to force an update instead of an insert when calling save() fotr the second time (in the other form). All of this apply if you are doing all of it in the same function on the same controller. Btw i'm just guessing.
On Jun 26, 2:19 am, francky06l <[EMAIL PROTECTED]> wrote: > you could use some javascript to hide or show the form when the user > checks/unchecks the box. > If you want the form appearing after submit you can use ajax to update > a div with your new form. > > On Jun 25, 10:40 pm, Mauro <[EMAIL PROTECTED]> wrote: > > > hi my name is Mauro, > > > i'm having a problem: > > > I have a checkbox in a page, if this checkbox is true another form has > > to be activated, how can i do that? > > > I tried calling (render) another view with the new form,once submit is > > push, but the info is saved in another registry on the table (mysql) > > > I don't know what to do > > > Thank you very much --~--~---------~--~----~------------~-------~--~----~ 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 -~----------~----~----~----~------~----~------~--~---
