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How display errormessage from model on jsp page?  RSS feed

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I plan write a login example with Netbean6.5 and Spring2.5.

I write a jsp page(login.jsp) to save user's account and password into command object,
then send to controller (LoginAction.java).

On LoginAction.java I return a model to login.jsp in order to display error message,
but I dont know how to display it on jsp page.

I read API, errors.getModel() is a model with hashmap class,
and I add two error messages by rejectValue(),
but What express language should I write on login.jsp,
then I can display the error messages on login.jsp?

This example is from book,but it's can't run,
so I fix the code.
Has anyone can tell me what should I do?

Document : login
Created on : 2009/4/3, 上午 09:55:47
Author : Administrator

<%@page contentType="text/html" pageEncoding="UTF-8"%>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"

<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title>JSP Page</title>


<spring:bind path="command.*">


<form name="loginform" method="post" action="validator.do">

<spring:bind path="command">

名稱<input type="text" name="account" value="${command.account}"/>

密碼<input type="password" name="password" value="${command.password}"/>


<input type="submit" value="送出"/>

<%--<vld:validate validationName="loginform" page="0"/>--%>


package edu.chenghao.internal.common.web.controller;

import java.util.ArrayList;
import java.util.HashMap;

import org.springframework.validation.BindException;
import org.springframework.web.servlet.ModelAndView;
import org.springframework.web.servlet.mvc.SimpleFormController;

import edu.chenghao.internal.common.web.form.LoginForm;
import java.util.List;

public class LoginAction extends SimpleFormController
public LoginAction()

protected ModelAndView onSubmit(Object arg0, BindException errors) throws Exception

LoginForm loginForm=(LoginForm)arg0;

if(loginForm.getAccount().equals("test") && loginForm.getPassword().equals("1234"))
HashMap hm=new HashMap();
ArrayList al=new ArrayList();
return new ModelAndView(this.getSuccessView(),hm);

List list = errors.getAllErrors();
for( int i=0; i<list.size(); i++){
System.out.println( "第"+i+"個"+ list.get(i) );


return new ModelAndView(this.getFormView(),errors.getModel());
//return new ModelAndView(this.getFormView(),this.getCommandName(),loginvo);

[Thumbnail for TestLoginForm.jpg]
(jpg to zip) This is my code.
Posts: 17314
IntelliJ IDE Mac Spring
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Well, I can't help you much with how you are doing it there. As I believe that is how you might have done it before Spring 2.5 Maybe Spring 2.0 or before.

But for me, I create a login page exactly like I would with straight JSP using the j_ properties in the login page.

Then in Spring, I would configure Spring Security and set a <login-page> to your page.

In your web.xml you use FORM for security, all this is from the Servlet/JSP spec, and it follows it to a tee.

I hope that helps move you in the right direction.

Good Luck

Cheng-Hao Hsu
Posts: 3
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Thanks your advise.

I need a job, so I learn Spring just because many companies project use it.
But you are right that use Servlet/jsp are basic and suitable.

Maybe I should learn the basic skills again.

Don't get me started about those stupid light bulbs.
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