Spring MVC一个简单的登录小程序如何跑不起来

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">


</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="prefix" value="/WEB-INF/jsp/"/>
<property name="suffix" value=".jsp"/>

</bean>

<bean id="urlMapping"
class="org.springframework.web.servlet.handler.SimpleUrlHandlerMapping">
<property name="mappings">
<props>
<prop key="login.do">LoginAction</prop>
</props>
</property>
</bean>

<bean id="LoginAction" class="cn.lxh.grth.action.LoginAction">
<property name="fail_view">
<value>Fail</value>
</property>

<property name="succeed_view">
<value>Succeed</value>
</property>

</bean>
</beans>

LoginAction代码

package cn.lxh.grth.action;

import java.net.BindException;
import java.util.logging.SimpleFormatter;

import org.springframework.web.servlet.ModelAndView;

import cn.lxh.grth.model.User;
public class LoginAction extends SimpleFormatter {

private String Succee_view="Succeed";

private String Fail_view="Fail";

public ModelAndView onSubmit(Object obj,BindException ex)throws Exception{
User u=(User)obj;
if("admin".equals(u.getName())&&"admin".equals(u.getPassword())){
return new ModelAndView(Succee_view);

}else{
return new ModelAndView(Fail_view);
}

}

public String getSuccee_view() {
return Succee_view;
}

public String getFail_view() {
return Fail_view;
}

public void setSuccee_view(String succeeView) {
Succee_view = succeeView;
}

public void setFail_view(String failView) {
Fail_view = failView;
}

}

User代码

package cn.lxh.grth.model;

public class User {

private String name;
private String password;
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public String getPassword() {
return password;
}
public void setPassword(String password) {
this.password = password;
}

}

[解决办法]
你难道连异常或者问题都不描述下吗谁会闲到看完你整个代码
[解决办法]
action=“login.do”直接一点

Spring MVC一个简单的登录小程序如何跑

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