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 <[email protected]> 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 <[email protected]> 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. >> >> --------------------------------------------------------------------- >> To unsubscribe, e-mail: [email protected] >> For additional commands, e-mail: [email protected] >> >> >
