IFeedbackProvider {
FeedbackPanel getFeedbackPanel();
}
MyPage extends Webpage implements IFeedbackProvider {
public MyPage() {
add(new FeedbackPanel("feedback"));
add(new MySubPanel("panel", this));
}
@Override
FeedbackPanel getFeedbackPanel() {
return get("feedback");
}
}
MySubPanel extends Panel {
private IFeedbackProvider feedback;
public MySubPanel(String id, IFeedbackProvider provider) {
super(id);
this.feedback = provider;
... some ajax handler ...
onSubmit(AjaxRequestTarget t) {
t.addComponent(feedback.getFeedbackPanel());
}
}
}
On Mon, Jan 18, 2010 at 5:52 PM, Martin Asenov <mase...@velti.com> wrote:
> Yes, but in my case the panels don't even know about the parent - they don't
> have a reference to it. I've got to think of some workaround on this one.
>
> Thanks for your time,
> Martin
>
> -----Original Message-----
> From: Jeremy Thomerson [mailto:jer...@wickettraining.com]
> Sent: Monday, January 18, 2010 6:45 PM
> To: users@wicket.apache.org
> Subject: Re: submit a form from outside of it
>
> Ah - I see. Yeah - you just have to roll your own option for this now. I'd
> just recommend some sort of listener pattern - something like adding a void
> formSubmitted(form, requesttarget) method on the page that child forms can
> call. Then the pages can do whatever they need with it.
>
> --
> Jeremy Thomerson
> http://www.wickettraining.com
>
>
>
> On Mon, Jan 18, 2010 at 10:40 AM, Martin Asenov <mase...@velti.com> wrote:
>
>> Yeah, Jeremy, that's pretty clear. I was saying that I can't know in the
>> parent page when the submit button is pressed , so that I can say
>> target.addComponent(feed); when feed is in parent page...
>>
>> BR,
>> Martin
>>
>> -----Original Message-----
>> From: Jeremy Thomerson [mailto:jer...@wickettraining.com]
>> Sent: Monday, January 18, 2010 6:37 PM
>> To: users@wicket.apache.org
>> Subject: Re: submit a form from outside of it
>>
>> What do you mean - you can't tell which button was pressed?
>>
>> Just add an onSubmit to the button and inside of it, add your feedback
>> message. or call getPage().info("..."), etc. Then allow the onSubmit of
>> the form to do its thing.
>>
>> --
>> Jeremy Thomerson
>> http://www.wickettraining.com
>>
>>
>>
>> On Mon, Jan 18, 2010 at 10:35 AM, Martin Asenov <mase...@velti.com> wrote:
>>
>> > Thank you both for the replies! After all I put the buttons in
>> myFormPanel.
>> > But this way I can't know when the submit and cancel buttons are pressed
>> so
>> > that I can render the feedback that is located in the parent page.
>> >
>> > Should I think of some listener?
>> >
>> > BR,
>> > Martin
>> >
>> > -----Original Message-----
>> > From: Jeremy Thomerson [mailto:jer...@wickettraining.com]
>> > Sent: Monday, January 18, 2010 6:31 PM
>> > To: users@wicket.apache.org
>> > Subject: Re: submit a form from outside of it
>> >
>> > Or wrap the outer page in a form so that any nested forms work with your
>> > out-of-place submit button.
>> >
>> > --
>> > Jeremy Thomerson
>> > http://www.wickettraining.com
>> >
>> >
>> >
>> > On Mon, Jan 18, 2010 at 9:43 AM, Alexandru Barbat <
>> > alexandrubar...@gmail.com
>> > > wrote:
>> >
>> > > I think you have to pass the form to the behavior in some way or you
>> can
>> > do
>> > > something like this..but it is ugly in some way :)
>> > >
>> > >
>> > > 1. in the form panel
>> > >
>> > > AjaxFormSubmitBehavior behave = new AjaxFormSubmitBehavior(form,
>> > > "onchange") {
>> > > ...
>> > > public void renderHead(IHeaderResponse response) {
>> > > super.renderHead(response);
>> > > response.renderJavascript("function submit_my_form(){\n"
>> +
>> > > getEventHandler().toString() + "\n}", "submit_my_form");
>> > > }
>> > > };
>> > >
>> > > form.add(behave);
>> > >
>> > >
>> > > ...
>> > >
>> > > 2. anywhere in the page
>> > >
>> > > and your button will look like this:
>> > >
>> > > <input type="button" value="my_button" onclick="submit_my_form()"/>
>> > >
>> > >
>> > >
>> > >
>> > > On Mon, Jan 18, 2010 at 12:22 PM, Martin Asenov <mase...@velti.com>
>> > wrote:
>> > >
>> > > > Hi, Alexandru, thanks for the quick reply.
>> > > >
>> > > > I get
>> > > >
>> > > > java.lang.IllegalStateException: form was not specified in the
>> > > constructor
>> > > > and cannot be found in the hierarchy of the component this behavior
>> is
>> > > > attached to
>> > > >
>> > > > the form is located in the same page and displayed, but it's actually
>> > > > placed within a panel that is a child of the page. :-(
>> > > >
>> > > > BR,
>> > > >
>> > > > -----Original Message-----
>> > > > From: Alexandru Barbat [mailto:alexandrubar...@gmail.com]
>> > > > Sent: Monday, January 18, 2010 11:26 AM
>> > > > To: users@wicket.apache.org
>> > > > Subject: Re: submit a form from outside of it
>> > > >
>> > > > Hi,
>> > > >
>> > > > Try this:
>> > > >
>> > > > AjaxFormSubmitBehavior behave = new AjaxFormSubmitBehavior(myForm,
>> > > > "onclick") {
>> > > > protected void onSubmit(AjaxRequestTarget target) {
>> > > > //do what you have to do
>> > > > }
>> > > >
>> > > >
>> > > > };
>> > > >
>> > > >
>> > > > Button submitButton = new Button("submitButton");
>> > > >
>> > > > submitButton.add(behave);
>> > > >
>> > > > ...
>> > > >
>> > > > Alexandru
>> > > >
>> > > > 2010/1/18 Martin Asenov <mase...@velti.com>
>> > > >
>> > > > > Hello, everyone!
>> > > > >
>> > > > > I have a form that has validation and so on, but the main
>> difference
>> > to
>> > > > > ordinary forms is that my form does not contain it's submit button.
>> > > It's
>> > > > > located in a parent, in my case a web page.
>> > > > >
>> > > > > I'm wondering how can I force the form submitting from the page.
>> The
>> > > code
>> > > > > is
>> > > > >
>> > > > > submitButton = new AjaxButton("submit_button) {
>> > > > > protected void onSubmit(AjaxRequestTarget target, Form<?>
>> > form)
>> > > {
>> > > > > myForm.processForm();
>> > > > > }
>> > > > > };
>> > > > >
>> > > > > The method processForm() in myForm just calls process();
>> > > > >
>> > > > > But nothing happens, looks like I'm missing something...
>> > > > >
>> > > > > Thanks,
>> > > > > Martin
>> > > > >
>> > > > >
>> > > >
>> > > > ---------------------------------------------------------------------
>> > > > To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org
>> > > > For additional commands, e-mail: users-h...@wicket.apache.org
>> > > >
>> > > >
>> > >
>> >
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>> >
>> >
>>
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>>
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