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From Sebastien <seb...@gmail.com>
Subject Re: Need help in implementing Ajax form
Date Mon, 14 May 2012 20:10:38 GMT
Damed! I just make me realizing I wrote "the anonymous method onSubmit"! I
cannot believe it...

On Mon, May 14, 2012 at 9:59 PM, W Mazur <wlodekmazur@gmail.com> wrote:

> Hi,
> -  Use Firefox with Firebug to track down client-server comunication
> (or lack of it) - Do you have javascript enebled? If terget == null
> then nothing happens in onSubmit() method.
> - Add some logging (log4j etc.), to your app
>
> Regards
> Wlodek
>
> 2012/5/14 Sebastien <sebfz1@gmail.com>:
> > And the WICKET AJAX DEBUG mark appears as soon as you are dealing with
> > wicket ajax component. It is not displayed anymore when your
> configuration
> > changes from "development" to "deployment" (web.xml)
> >
> > On Mon, May 14, 2012 at 9:47 PM, Sebastien <sebfz1@gmail.com> wrote:
> >
> >> Hi kshitiz,
> >>
> >> Well, looking at the code as-is does not make me understand what is
> going
> >> wrong. I mean, it should go until onSubmit(). For you implementation, I
> can
> >> just notice that you did not reattach any feedback panel to the target.
> >> Maybe you do not see the error (but if it happens, you can see it in the
> >> console as "Component feedback message was left unrendered..."); and you
> >> changed the reference to the userDomain object. the model is then not
> >> sync. You need to do something like
> loginForm.setModelObject(userDomain).
> >> Mark the form as final to be able to access it from the anonymous method
> >> onSubmit.
> >>
> >> But apart from that, I just would like to tell you that if you need an
> >> authentication mechanism, you'll probably better have to use the
> >> wicket-auth-roles.
> >> All you need to know is here:
> >> http://wicket.apache.org/learn/projects/authroles.html
> >>
> >> Regards,
> >> Sebastien
> >>
> >> On Mon, May 14, 2012 at 8:10 PM, kshitiz <k.agarwal4@gmail.com> wrote:
> >>
> >>> Hi,
> >>>
> >>> I am trying to learn Ajax in Wicket and implement it in my project. I
> am
> >>> trying to use AjaxFallbackButton but there is something I am missing.
> Here
> >>> is my code:
> >>>
> >>> StatelessForm<UserDomain> loginForm = new StatelessForm<UserDomain>(
> >>>                                "loginForm", new
> >>> CompoundPropertyModel<UserDomain>(userDomain));
> >>>
> >>>
> >>> AjaxFallbackButton ajaxSubmitButton = new AjaxFallbackButton(
> >>>                                "ajaxSubmitButton", loginForm) {
> >>>
> >>>                        /**
> >>>                         *
> >>>                         */
> >>>                        private static final long serialVersionUID = 1L;
> >>>
> >>>                        @Override
> >>>                        protected void onSubmit(AjaxRequestTarget
> target,
> >>> Form<?> form) {
> >>>
> >>>                                if(target!=null){
> >>>                                UserService userService = new
> >>> UserService();
> >>>                                try {
> >>>                                        userDomain =
> >>> userService.login(emailIdTextField.getValue(),
> >>>
> >>>  passwordTextField.getValue());
> >>>                                } catch (Exception exception) {
> >>>                                        error(exception.getMessage());
> >>>                                        error = true;
> >>>                                }
> >>>
> >>>                                if (!error) {
> >>>                                        setResponsePage(Home.class,
> >>> pageParameters);
> >>>                                }
> >>>                                }
> >>>                        }
> >>>
> >>>                        @Override
> >>>                        protected void onError(AjaxRequestTarget target,
> >>> Form<?> form) {
> >>>                                // TODO Auto-generated method stub
> >>>
> >>>                        }
> >>>                };
> >>>                loginForm.add(ajaxSubmitButton);
> >>> }
> >>>
> >>> Html is
> >>> <input type="submit" value="Login" class="formbutton" name="Login"
> >>> wicket:id="ajaxSubmitButton"/>
> >>>
> >>> But when I press Login button, nothing happens...Please help me out
> here
> >>> as
> >>> I am new to this. Is there any good doc available for this? Also, why
> >>> WICKET
> >>> AJAX DEBUG mark is coming in bottom right corner of my web page?
> >>>
> >>> --
> >>> View this message in context:
> >>>
> http://apache-wicket.1842946.n4.nabble.com/Need-help-in-implementing-Ajax-form-tp4632760.html
> >>> Sent from the Users forum mailing list archive at Nabble.com.
> >>>
> >>> ---------------------------------------------------------------------
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> >>> For additional commands, e-mail: users-help@wicket.apache.org
> >>>
> >>>
> >>
>
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>

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