Spring MVC 入门，从简单登陆做起_sts做一个简单的登录mvc-优快云博客

本文链接：https://blog.youkuaiyun.com/sust2012/article/details/7736676

本文详细介绍了如何使用Spring框架实现一个简单的登录系统，包括配置web.xml、映射文件、控制器类、登录页面和成功页面。

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第一步在 web.xml中配置如下：

<?xml version="1.0" encoding="UTF-8"?>

<web-app version="2.4" xmlns="http://java.sun.com/xml/ns/j2ee"

xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"

xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee

http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">

<context-param>

<param-name>contextConfigLocation</param-name>

<param-value>/WEB-INF/applicationContext.xml</param-value>

</context-param>

<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>

</listener>

<filter-name>encodingFilter</filter-name>

<filter-class>org.springframework.web.filter.CharacterEncodingFilter</filter-class>

<init-param>

<param-name>encoding</param-name>

<param-value>UTF-8</param-value>

</init-param>

</filter>

<filter-mapping>

<filter-name>encodingFilter</filter-name>

<url-pattern>*.do</url-pattern>

</filter-mapping>

<servlet-name>login</servlet-name>

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

<load-on-startup>1</load-on-startup>

</servlet>

<servlet-mapping>

<servlet-name>login</servlet-name>

<url-pattern>*.do</url-pattern>

</servlet-mapping>

</web-app>

第二部配置映射文件，此映射文件的文件名为web.xml中配置的DispatcherServlet的名字-servlet.xml，我的映射文件为login-servlet.xml,此配置文件将会被web容器自动加载，内容如下：

<?xml version="1.0" encoding="UTF-8"?>

<!DOCTYPE beans PUBLIC "-//SPRING//DTD BEAN//EN"

"http://www.springframework.org/dtd/spring-beans.dtd">

<beans>

</bean>

<props>

<prop key="/login.do">LoginController</prop>

</props>

</property>

</bean>

</beans>

其中 prefix表示存放jsp文件的位置，suffix指定为jsp文件

第三步进入Controller

public class LoginController implements Controller{

public ModelAndView handleRequest(HttpServletRequest request,

HttpServletResponse response) throws Exception {

String userName = request.getParameter("userName");

String userPass = request.getParameter("userPass");

if("devin".equals(userName)&&"jintaiyang".equals(userPass)){

request.setAttribute("userName", userName);

return new ModelAndView("success");

}

else{

request.setAttribute("error","用户名和密码错误");

return new ModelAndView("login");

}

上面的success和login都是逻辑名，只需保证success.jsp和login.jsp存在就行

第四步 login.jsp

<%@ page language="java" contentType="text/html; charset=utf-8"
pageEncoding="utf-8"%>
<!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>Insert title here</title>
</head>
<body>
<form action="login.do" method="post">
userName:<input type="text" name="userName"><br>
userPass:<input type="password" name="userPass"><br>
<input type="submit" value="登陆"><input type="reset" value="reset" value="重置"><br>
<font color="red">${error }</font>
</form>
</body>
</html>