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!
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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 instances of type MyDialog to be re-rendered
in this Ajax request/response.
Then in your MyDialog#onBeforeRender() { target = AjaxRequestTarget.get();
doSomethingWithTarget(target); }

On Tue, Dec 14, 2010 at 9:16 PM, guydog28 g...@guydo.com wrote:


 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 it or add components.  I'm really hoping someone has a
 suggestion that I missed.  This is a major blocker for us right now.
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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 MyAjaxSubmitLink extends AjaxSubmitLink {

private static final long serialVersionUID = 1L;

/**
 * @param id
 */
public MyAjaxSubmitLink(String id) {
super(id);
}

/**
 * @param id
 * @param form
 */
public MyAjaxSubmitLink(String id, Form? form) {
super(id, form);
}

@Override
protected final void onSubmit(AjaxRequestTarget target, Form? form) {
IAjaxSubmitAware submitAware = 
findParent(IAjaxSubmitAware.class);
if(submitAware != null) {
submitAware.onSubmit(target);
}
doOnSubmit(target, form);
}

protected abstract void  doOnSubmit(AjaxRequestTarget target, Form? 
form);
}

Ernesto


On Tue, Dec 14, 2010 at 9:16 PM, guydog28 g...@guydo.com wrote:

 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 it or add components.  I'm really hoping someone has a
 suggestion that I missed.  This is a major blocker for us right now.
 --
 View this message in context: 
 http://apache-wicket.1842946.n4.nabble.com/How-to-listen-for-CHILD-ajax-form-submits-tp3086439p3087882.html
 Sent from the Users forum mailing list archive at Nabble.com.

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 For additional commands, e-mail: users-h...@wicket.apache.org



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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 it or add components.  I'm really hoping someone has a
suggestion that I missed.  This is a major blocker for us right now.
-- 
View this message in context: 
http://apache-wicket.1842946.n4.nabble.com/How-to-listen-for-CHILD-ajax-form-submits-tp3086439p3087882.html
Sent from the Users forum mailing list archive at Nabble.com.

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