views:

2084

answers:

3

I am using Spring Web MVC and Hibernate for developing my application.

My login.jsp page has following code :

<form:form method="post" commandName="User">
   User Name : 
      <form:input path="email"/>
   Password : 
     <form:input path="password"/>

<input type="submit" align="center" value="Execute">

Now, My servlet.xml file has following code :

 <bean name="/uservalidate.htm" class="com.sufalam.mailserver.presentation.web.UserValidateFormController">
        <property name="sessionForm" value="true"/>
        <property name="commandName" value="User"/>
        <property name="commandClass" value="com.sufalam.mailserver.bean.User"/>
        <property name="formView" value="login"/>
        <property name="successView" value="layout.jsp"/>
        <property name="userSecurityProcessor" ref="IUserSecurityProcessor"/>
    </bean>

My UserValidateFormController has following code :

public class UserValidateFormController extends SimpleFormController {

    /** Logger for this class and subclasses */
    protected final Log logger = LogFactory.getLog(getClass());
    private IUserSecurityProcessor userSecurityProcessor;

    public ModelAndView onSubmit(Object command)
            throws ServletException, SufalamException {
            ModelAndView mav = new ModelAndView();
            Map model = new HashMap();


        String username = ((User) command).getEmail();
        String password = ((User) command).getPassword();
        List userChecking = new ArrayList();
        userChecking = userSecurityProcessor.findByAll(username, password, 0.0);
        System.out.println("userChecking length = "+userChecking.size());
        if (userChecking.size() == 1) {
            return new ModelAndView("layout");
            //return new ModelAndView(new RedirectView(getSuccessView()));
        }

        return new ModelAndView("login", model);

    }

    protected Object formBackingObject(HttpServletRequest request) throws ServletException {
        User user = new User();
        return user;
    }

  public void setUserSecurityProcessor(IUserSecurityProcessor userSecurityProcessor) {
        this.userSecurityProcessor = userSecurityProcessor;

    }

In my UserValidateFormController at the time of handing submit event, i am checking that username and password are correct or not..

It's working fine & if both are matching then its redirecting to layout.jsp, that's also fine.

But if username or password are incorrect then i want to redirect to same login.jsp page and display appropriate error..

Please suggest me the solution that what to do redirecting to same view controller..

Thanks in advance..

A: 

... not sure if this is what you're looking for, but is this what you're looking for:

    else { return new ModelAndView( "login", model ); }

... otherwise I missed something in your question. It seems to me you've pretty far to get stuck like this.

vector
I can't do that, as my login page requires User to be initialized... So, it will be in trouble and give me an error of can't find User Command from login.jsp..
Nirmal
... then what about: return new ModelAndView( new RedirectView ( getFormView() ), model );
vector
A: 

I would say all you have to do is populate your model before using it again:

    if (userChecking.size() == 1) {
        return new ModelAndView("layout");
        //return new ModelAndView(new RedirectView(getSuccessView()));
    }
    model.put("User", command);
    return new ModelAndView("login", model);
Triqui
A: 

Finally solve this issue with following line of code :

return new ModelAndView(new RedirectView(getSuccessView()));

or

return new ModelAndView(new RedirectView("success.htm");

Thanks...

Nirmal