).
Thank you all!
Best Regards,
Martin
-Original Message-
From: Martijn Dashorst [mailto:martijn.dasho...@gmail.com]
Sent: Monday, January 18, 2010 7:12 PM
To: users@wicket.apache.org
Subject: Re: submit a form from outside of it
IFeedbackProvider {
FeedbackPanel getFeedbackPanel
me 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
&g
day, 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) met
n 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
aining.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(&qu
e.
>
> 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
>
ssage-
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
:-(
> >
> > 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
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") {
>prot
d to solve this.
Please, if someone has a better idea, it would be great to share it...
Best Regards,
Martin
-Original Message-
From: Martin Asenov [mailto:mase...@velti.com]
Sent: Monday, January 18, 2010 3:33 PM
To: users@wicket.apache.org
Subject: RE: submit a form from outside of
g it persistent.
Thanks,
-Original Message-
From: Bert [mailto:taser...@gmail.com]
Sent: Monday, January 18, 2010 2:17 PM
To: users@wicket.apache.org
Subject: Re: submit a form from outside of it
I 'm not sure i completely understand your requirement, but the
AjaxButton too has a con
I 'm not sure i completely understand your requirement, but the
AjaxButton too has a constructor
with a Form as parameter. Would it be possible to access the form when
creating the button?
Bert
On Mon, Jan 18, 2010 at 13:11, Martin Asenov wrote:
> Thanks Bert, but I use a button... I need to hav
t: Re: submit a form from outside of it
Not sure here, but the AjaxSubmitLink has an constructor that lets you
pass in the form it should work on.
Bert
-
To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org
For additional comman
Not sure here, but the AjaxSubmitLink has an constructor that lets you
pass in the form it should work on.
Bert
-
To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org
For additional commands, e-mail: users-h...@wicket.apac
hin 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
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(b
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
submitButto
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