Re: How to listen for CHILD ajax form submits

2010-12-16 Thread guydog28
Ernesto, I had thought of that. I was hoping to avoid the custom button/link route, but seeing as I haven't heard anything more elegant yet, I decided to go with it for now. Thanks for your input! -- View this message in context:

Re: How to listen for CHILD ajax form submits

2010-12-15 Thread Martin Grigorov
if you use 1.5 then you can use the new event mechanism. see WICKET-1312 in 1.4 you can use something like: new AjaxFormSubmitBehavior(event) { onSubmit(AjaxRequestTarget target, Form form) { ... target.addChildren(getPage(), MyDialog.class); } } The last line will register all

Re: How to listen for CHILD ajax form submits

2010-12-15 Thread Ernesto Reinaldo Barreiro
If you have full control over the contents of your dialog you could try to 1- create the interface to mark your dialog: public interface IAjaxSubmitAware { void onSubmit(AjaxRequestTarget target); } 2- create your own link, buttons, etc as in public abstract class

Re: How to listen for CHILD ajax form submits

2010-12-14 Thread guydog28
Anyone? Abstract: I have extended WiQuery Dialog with our own custom dialog. I want the MyDialog to know if any child components are a Form, being submitted via AJAX (AjaxFormSubmitBehavior). I need access to the AjaxRequestTarget created by that ajax form submit so I can append javascript to