AjaxFallbackButton resetting validation status of form?

[Mike Comb <mc...@softcoin.com>]

I've got a form that is a simple collection of TextFields.  I'd like  
to validate that form through AJAX prior to submission and redraw the  
TextFields which have a Behavior attached to them that changes the  
font color if they are invalid.  To do that I've added an  
AjaxFallbackButton that looks like this:

		AjaxFallbackButton button = new AjaxFallbackButton (id, form) {
			@Override
			protected void onSubmit(AjaxRequestTarget target, Form form) { }
			
			@Override
			protected void onError(final AjaxRequestTarget target, Form form) {
				// Redraw the feedback panel
				if(feedbackPanel!=null)
					target.addComponent(feedbackPanel);
				
				// Redraw all child textfields
				target.addChildren(form,TextField.class);
			}
		};

This works perfectly the first time it is clicked.  However, if the  
user does not correct the error or if there is a second error no  
validation occurs on the second submission.  If I comment out the  
"addChildren" call the form works as I would expect and all errors  
are shown in the FeedbackPanel through multiple submissions.  Doing  
anything that visits the form components seems to cause them to show  
as valid the next time the form is submitted (I've also tried using  
Form.visitFormComponents() and MarkupContainer.iterator() with  
similar results).

Do I need to do something to reset the state of the form after  
visiting the child components or could this be a bug?  Is there a  
better way to do this that I'm missing?  It happens in both 1.3b3 and  
1.3b4.

Note that I can't use AjaxFormValidatingBehavior because I need all  
the validation to happen in one pass when the form is submitted not  
as the user types things in.

Thanks,
-Mike
--
Mike Comb
Director of Engineering
SoftCoin, Inc
mcomb@softcoin.com



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