请教：一直404错误，到底应该怎么改路径？

baidu_35892047 2016-08-31 03:26:17

完全是copy网上的例子

web.xml
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://java.sun.com/xml/ns/javaee" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" id="WebApp_ID" version="3.0">
<display-name>springmvc-1</display-name>
<welcome-file-list>
<welcome-file>index.html</welcome-file>
<welcome-file>index.htm</welcome-file>
<welcome-file>index.jsp</welcome-file>
<welcome-file>default.html</welcome-file>
<welcome-file>default.htm</welcome-file>
<welcome-file>default.jsp</welcome-file>
</welcome-file-list>

<servlet>
<servlet-name>spring</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
</servlet>

<servlet-mapping>
<servlet-name>spring</servlet-name>
<url-pattern>*.do</url-pattern>
</servlet-mapping>
</web-app>

TestController.java
package com.itcount.springmvc;

import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

import org.springframework.web.servlet.ModelAndView;
import org.springframework.web.servlet.mvc.AbstractController;

public class TestController extends AbstractController {

@Override
protected ModelAndView handleRequestInternal(HttpServletRequest arg0,
HttpServletResponse arg1) throws Exception {
// TODO Auto-generated method stub
System.out.println("hello springmvc");
return new ModelAndView("index.jsp");
}

}

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"

xmlns:aop="http://www.springframework.org/schema/aop"
xmlns:tx="http://www.springframework.org/schema/tx"
xsi:schemaLocation="http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
http://www.springframework.org/schema/context
http://www.springframework.org/schema/context/spring-context-3.0.xsd

http://www.springframework.org/schema/mvc
http://www.springframework.org/schema/mvc/spring-mvc-3.0.xsd
http://www.springframework.org/schema/aop
http://www.springframework.org/schema/aop/spring-aop-3.0.xsd
http://www.springframework.org/schema/tx
http://www.springframework.org/schema/tx/spring-tx-3.0.xsd">

<bean id="testController" name="/hello.do" class="com.itcount.springmvc.TestController"></bean>

<bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<property name="prefix" value="/WEB-INF/jsp/"></property>
<property name="suffix" value=".jsp"></property>
</bean>

</beans>

index.jsp
<%@ page language="java" contentType="text/html; charset=utf-8"%>
<%
String path = request.getContextPath();
String basePath = request.getScheme()+"://"+request.getServerName()+":"+request.getServerPort()+path+"/";
%>

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
<head>
<base href="<%=basePath%>">

<title>My JSP 'index.jsp' starting page</title>
<meta http-equiv="pragma" content="no-cache">
<meta http-equiv="cache-control" content="no-cache">
<meta http-equiv="expires" content="0">
<meta http-equiv="keywords" content="keyword1,keyword2,keyword3">
<meta http-equiv="description" content="This is my page">

</head>

<body>
This is my JSP page. <br>
</body>
</html>

我知道404是路径错误，但是不知道怎么改，请教

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tangwwip 2016-09-09

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你这项目不完整呀，少下了东西吧

李大白白 2016-09-09

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上图可以看出来, 你的后缀已经写了.jsp, 那么返回modelAndView的时候不需要加.jsp了

还有,请问你是访问index.jsp就404吗? 你的url是什么, index..jsp是放在WEB-INF文件夹里?

Swen程序员 2016-09-09

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~~~每个项目启动都是先访问配置文件的，你这个在tomcat启动估计都报错了。

书生_iit 2016-09-08

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估计不完整, 看看xml文件一步一步找下去

Gsprinkle 2016-09-03

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<servlet> <servlet-name>DispatcherServlet</servlet-name> <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>  <init-param> <param-name>contextConfigLocation</param-name> <param-value>classpath:spring.xml</param-value> </init-param> </servlet> <servlet-mapping> <servlet-name>DispatcherServlet</servlet-name> <url-pattern>*.action</url-pattern> </servlet-mapping> 给你一个例子参考一下

Gsprinkle 2016-09-03