开发者社区> 问答> 正文

Spring MVC 无法访问.do文件

初学Spring MVC,照着template写了个简单的demo,发现.jsp文件可以正常访问,但是.do不知道为什么就报404了screenshot

HelloController.java文件如下:

import org.springframework.web.servlet.ModelAndView;
import org.springframework.web.servlet.mvc.Controller;
 
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
 
public class HelloController implements Controller{
    private String viewPage;
 
    @Override
    public ModelAndView handleRequest(HttpServletRequest httpServletRequest, HttpServletResponse httpServletResponse) throws Exception {
        String user = httpServletRequest.getParameter("user");
        return new ModelAndView(viewPage, "user", user);
    }
 
    public void setViewPage(String viewPage) {
        this.viewPage = viewPage;
    }
}
mvc-config.xml文件如下:

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
       xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
       xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd">
 
    <bean id="viewResolver" class="org.springframework.web.servlet.view.InternalResourceViewResolver">
        <property name="prefix" value="/"/>
        <property name="suffix" value=".jsp"/>
    </bean>
 
    <bean name="/hello.do" class="HelloController">
        <property name="viewPage" value="hello"/>
    </bean>
</beans>
web.xml文件如下:

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns="http://xmlns.jcp.org/xml/ns/javaee"
         xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
         xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd"
         version="3.1">
 
    <session-config>
        <session-timeout>30</session-timeout>
    </session-config>
 
    <servlet>
        <servlet-name>dispatcherServlet</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <init-param>
            <param-name>contextConfigLocation</param-name>
            <param-value>/WEB-INF/mvc-config.xml</param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>
     
    <servlet-mapping>
        <servlet-name>dispatcherServlet</servlet-name>
        <url-pattern>*.do</url-pattern>
    </servlet-mapping>
</web-app>
hello.jsp文件如下:


<%@ page contentType="text/html;charset=UTF-8" language="java" %>
<html>
  <head>
    <title>Spring MVC</title>
  </head>
  <body>
        <h1>Hello, ${user}!!</h1>
  </body>
</html>


服务器是tomcat 8.0.18,访问.jsp时的URL是 http://localhost:8080/hello.jsp

换成http://localhost:8080/hello.do就404了,试了下http://localhost:8080/Spring MVC/hello.do?user=wedny也是不行

请问各位大神知道是什么问题吗?

展开
收起
a123456678 2016-03-17 16:34:29 3522 0
1 条回答
写回答
取消 提交回答
  • 我试了下你的方法,在上面的HelloController.java改动如下:

    import org.springframework.web.bind.annotation.RequestMapping;
    import org.springframework.web.servlet.ModelAndView;
    import org.springframework.web.servlet.mvc.Controller;
     
    import javax.servlet.http.HttpServletRequest;
    import javax.servlet.http.HttpServletResponse;
     
    @RequestMapping("/test")
    public class HelloController implements Controller{
        private String viewPage;
     
        @RequestMapping("/first")
        @Override
        public ModelAndView handleRequest(HttpServletRequest httpServletRequest, HttpServletResponse httpServletResponse) throws Exception {
            String user = httpServletRequest.getParameter("user");
            return new ModelAndView(viewPage, "user", user);
        }
     
        public void setViewPage(String viewPage) {
            this.viewPage = viewPage;
        }
    }
    
    2019-07-17 19:05:38
    赞同 展开评论 打赏
问答排行榜
最热
最新

相关电子书

更多
云栖社区特邀专家徐雷Java Spring Boot开发实战系列课程(第20讲):经典面试题与阿里等名企内部招聘求职面试技巧 立即下载
微服务架构模式与原理Spring Cloud开发实战 立即下载
阿里特邀专家徐雷Java Spring Boot开发实战系列课程(第18讲):制作Java Docker镜像与推送到DockerHub和阿里云Docker仓库 立即下载

相关实验场景

更多