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[0528]2-springMVC入门准备工作二

2012-08-22 
[0528]2-springMVC入门准备工作21.修改web.xml,加入spring的请求控制器!-- 前端控制器 --servletserv

[0528]2-springMVC入门准备工作2

1.修改web.xml,加入spring的请求控制器

<!-- 前端控制器 -->  <servlet>  <servlet-name>dispatcher</servlet-name>  <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>  <!--<init-param>  <param-name>contextConfigLocation</param-name>  <param-value></param-value>  </init-param>  --></servlet>  <servlet-mapping>  <servlet-name>dispatcher</servlet-name>  <url-pattern>*.htm</url-pattern>  </servlet-mapping>

?2.根据上面的servlet-name,默认的配置bean文件:

dispatcher-servlet.xml

<bean id="viewResolver" value="/WEB-INF/jsp/" /><property name="suffix" value=".jsp" /></bean><bean name="/show.htm" />

?3.请求分发到控制LoginController

package com.iteye.niewj.controller;import javax.servlet.http.HttpServletRequest;import javax.servlet.http.HttpServletResponse;import org.springframework.web.servlet.ModelAndView;import org.springframework.web.servlet.mvc.Controller;import com.iteye.niewj.domain.User;import com.iteye.niewj.service.UserService;public class LoginController implements Controller {private UserService<User> userService; public void setUserService(UserService<User> userService) {this.userService = userService;}@Overridepublic ModelAndView handleRequest(HttpServletRequest request,HttpServletResponse response) throws Exception {String userName = request.getParameter("userName");//String passwd = request.getParameter("passwd");//User user = (User)userService.findUser(userName, passwd);return new ModelAndView("show", "user", userName);}}
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