Spring MVC一个简单的登录小程序如何跑不起来
日期：2014-05-19 浏览次数：20591 次
Spring MVC一个简单的登录小程序怎么跑不起来
用Spring MVC写的一个简单程序 可是到了登录页面提交就提示404错误 说找不到
HTTP Status 404 - /Login_System_01/login.do
代码虽然多 核心代码却很少的 希望大家可以帮我看看 我困惑了很久~~
代码如下~~
(Long.jsp)
<%@ page language="java" import="java.util.*" pageEncoding="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 'Login.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">
<!--
<link rel="stylesheet" type="text/css" href="styles.css">
-->
</head>
<body>
<form method="POST" action="/Login_System_01/login.do">
<p align="center">登录</p>
<br>
用户名:
<input type="text" name="username" >
<br>
密 码 :
<input type="password" name="password" >
<br>
<p>
<input type="submit" value="提交" name="B1">
<input type="reset" value="重置" name="B2">
</p>
</form>
</body>
</html>
对于失败页面 成功页面的代码就不发了~~~
然后web.xml代码
<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.5"
xmlns="http://java.sun.com/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee
http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">
<welcome-file-list>
<welcome-file>WEB-INF/jsp/Login.jsp</welcome-file>
</welcome-file-list>
<servlet>
<servlet-name>Dispatcher</servlet-name>
<servlet-class>
org.springframework.web.servlet.DispatcherServlet
</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Dispatcher</servlet-name>
<url-pattern>*.do</url-pattern>
</servlet-mapping>
</web-app>
dispatcherServlet-servlet.xml代码
<?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-2.5.xsd">
<bean id="viewResolver"
class="org.springframework.web.servlet.view.UrlBasedViewResolver">
<property name="viewClass" value="org.springframework.web.servlet.view.JstlView"/>
<property name=&quo
用Spring MVC写的一个简单程序 可是到了登录页面提交就提示404错误 说找不到
HTTP Status 404 - /Login_System_01/login.do
代码虽然多 核心代码却很少的 希望大家可以帮我看看 我困惑了很久~~
代码如下~~
(Long.jsp)
<%@ page language="java" import="java.util.*" pageEncoding="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 'Login.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">
<!--
<link rel="stylesheet" type="text/css" href="styles.css">
-->
</head>
<body>
<form method="POST" action="/Login_System_01/login.do">
<p align="center">登录</p>
<br>
用户名:
<input type="text" name="username" >
<br>
密 码 :
<input type="password" name="password" >
<br>
<p>
<input type="submit" value="提交" name="B1">
<input type="reset" value="重置" name="B2">
</p>
</form>
</body>
</html>
对于失败页面 成功页面的代码就不发了~~~
然后web.xml代码
<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.5"
xmlns="http://java.sun.com/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee
http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">
<welcome-file-list>
<welcome-file>WEB-INF/jsp/Login.jsp</welcome-file>
</welcome-file-list>
<servlet>
<servlet-name>Dispatcher</servlet-name>
<servlet-class>
org.springframework.web.servlet.DispatcherServlet
</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Dispatcher</servlet-name>
<url-pattern>*.do</url-pattern>
</servlet-mapping>
</web-app>
dispatcherServlet-servlet.xml代码
<?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-2.5.xsd">
<bean id="viewResolver"
class="org.springframework.web.servlet.view.UrlBasedViewResolver">
<property name="viewClass" value="org.springframework.web.servlet.view.JstlView"/>
<property name=&quo
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