Hi Marek,
you could override form method 'process(IFormSubmittingComponent
submittingComponent) ' checking whether submittingComponent is
submitButton or ajaxButton. submitButton follows standard behavior
calling super.process(...), whereas ajaxButton just call validate().
But remember that form's onSubmit() must be called anyway, so as Pedro
suggested you should move onSubmit code to submitButton.
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
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