3

我想按照示例实现spring3.2.0 web mvc：http: //programmersplanet.wordpress.com/2011/11/26/4/

但我失败了。最后我得到的结果是：Successfully logged in: ${user.username}，你看到变量是无用的（JSP 不起作用）。我添加了一些输出来调试，它证明了@Controller、@RequestMapping 工作。所以我不知道我的编码有什么问题。

spring-servlet.xml

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
       xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
       xmlns:context="http://www.springframework.org/schema/context"
       xmlns:mvc="http://www.springframework.org/schema/mvc"
       xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd
        http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context.xsd
         http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc.xsd">

    <context:annotation-config/>
    <mvc:annotation-driven/>
    <context:component-scan base-package="demo.spring"/>
    <bean id="viewResolver" class="org.springframework.web.servlet.view.InternalResourceViewResolver">
        <property name="prefix" value="/WEB-INF/views/"/>
        <property name="suffix" value=".jsp"/>
    </bean>
</beans>

登录控制器.java

import org.springframework.stereotype.Controller;
import org.springframework.validation.BindingResult;
import org.springframework.web.bind.annotation.ModelAttribute;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestMethod;
import org.springframework.web.bind.support.SessionStatus;
import org.springframework.web.servlet.ModelAndView;

@Controller
@RequestMapping(value = "/")
public class LoginController {
    @RequestMapping(method = RequestMethod.GET)
    public ModelAndView displayLoginView() {
        UserForm userForm = new UserForm();
        ModelAndView view = new ModelAndView("login");
        view.addObject(userForm);
        return view;
    }

    @RequestMapping(method = RequestMethod.POST)
    public ModelAndView doLogin(@ModelAttribute("user") UserForm user, BindingResult bindingResult, SessionStatus sessionStatus) {
        ModelAndView model = new ModelAndView("home");
        System.out.println(user.getUsername());// this line is right.
        return model;
    }
}

主页.jsp

<%@ page contentType="text/html;charset=UTF-8" language="java" %>

<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
    <meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
    <title>home</title>
</head>
<body>
Successfully logged in: ${user.username}
</body>
</html>

web.xml

<!DOCTYPE web-app PUBLIC
        "-//Sun Microsystems, Inc.//DTD Web Application 2.3//EN"
        "http://java.sun.com/dtd/web-app_2_3.dtd" >

<web-app>
    <display-name>Archetype Created Web Application</display-name>
    <servlet>
        <servlet-name>DispatchServlet</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <init-param>
            <param-name>contextConfigLocation</param-name>
            <param-value>/WEB-INF/spring-servlet.xml</param-value>
        </init-param>
    </servlet>
    <servlet-mapping>
        <servlet-name>DispatchServlet</servlet-name>
        <url-pattern>/</url-pattern>
    </servlet-mapping>
</web-app
4

2 回答 2

2

您的问题的可能原因是您在控制器中添加了“userForm”并尝试在 JSP 中以 ${user.username} 的形式访问它！

还要检查您是否在 web.xml 中禁用了 EL 表达。我们有 1000 名用户依赖 Spring MVC，这是迄今为止最可靠的 Java EE 框架。

于 2012-12-19T06:00:00.897 回答
0

我的 web.xml 是由 maven 自动生成的。这是错误的。

修改前（默认）：

<!DOCTYPE web-app PUBLIC "-//Sun Microsystems, Inc.//DTD Web Application 2.3//EN" "http://java.sun.com/dtd/web-app_2_3.dtd" >

修改后：

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" version="2.5" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee   http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">
于 2012-12-19T06:16:37.660 回答