An idea: in Form#onSubmit() check whether the current request is Ajax: if (AjaxRequestTarget.get() != null) return; // i.e. do not execute the method' body
You can even check whether the submittingComponent is the AjaxButton and do nothing in that case only On Wed, Jan 5, 2011 at 1:19 PM, Marek Pribyl <mr.pri...@gmail.com> wrote: > hi all, > I have form with two submit buttons & want different behavior for each > button. > > submitButton: > - standard submit button > - expected "onClick" behavior: validate form & save the form's model > (implemented in form's onSubmit() method) & redirect to another page > > ajaxButton: > - AjaxButton > - expected "onClick" behavior: validate form, if valid get form's model > & display it as XML in modal window for preview > > the problem is that form's onSubmit() is called when I click on > ajaxButton - I need to keep validation but do not call form's onSubmit() > method, i.e. stay on the current form page & do not save the model. > Is there any chance to do this wicket way? > > thanks, > marek > > > --------------------------------------------------------------------- > To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org > For additional commands, e-mail: users-h...@wicket.apache.org > >
