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 :
> 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 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 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 loginForm = new StatelessForm(
>>> "loginForm", new
>>> CompoundPropertyModel(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
>>> >> 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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>>>
>>
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