以下示例说明如何使用Spring Web MVC框架使用Multi Action Controller的Parameterizable View Controller方法.可参数化视图允许使用请求映射网页.
package com.it1352;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import org.springframework.web.servlet.ModelAndView;
import org.springframework.web.servlet.mvc.multiaction.MultiActionController;
public class UserController extends MultiActionController{
public ModelAndView home(HttpServletRequest request,
HttpServletResponse response) throws Exception {
ModelAndView model = new ModelAndView("user");
model.addObject("message", "Home");
return model;
}
}<bean class="org.springframework.web.servlet.handler.SimpleUrlHandlerMapping"> <property name="mappings"> <value> index.htm=userController </value> </property> </bean> <bean id="userController" class="org.springframework.web.servlet.mvc.ParameterizableViewController"> <property name="viewName" value="user"/> </bean>
例如，使用上述配置，如果是URI.
首先，让我们使用一个可用的Eclipse IDE，并坚持以下步骤，使用Spring Web Framework开发基于动态表单的Web应用程序.
| Step | Description |
|---|---|
| 1 | 创建一个名为TestWeb的项目打包com.it1352，如Spring MVC中所述 - Hello World章节. |
| 2 | 在com.it1352包下创建一个Java类UserController. |
| 3 | 在jsp子文件夹下创建一个视图文件user.jsp. |
| 4 | 最后一步是创建源文件和配置文件的内容并导出申请如下所述. |
package com.it1352;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import org.springframework.web.servlet.ModelAndView;
import org.springframework.web.servlet.mvc.multiaction.MultiActionController;
public class UserController extends MultiActionController{
public ModelAndView home(HttpServletRequest request,
HttpServletResponse response) throws Exception {
ModelAndView model = new ModelAndView("user");
model.addObject("message", "Home");
return model;
}
}<beans xmlns = "http://www.springframework.org/schema/beans" xmlns:context = "http://www.springframework.org/schema/context" 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-3.0.xsd http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-3.0.xsd"> <bean class = "org.springframework.web.servlet.view.InternalResourceViewResolver"> <property name = "prefix" value = "/WEB-INF/jsp/"/> <property name = "suffix" value = ".jsp"/> </bean> <bean class = "org.springframework.web.servlet.handler.SimpleUrlHandlerMapping"> <property name = "mappings"> <value> index.htm = userController </value> </property> </bean> <bean id = "userController" class = "org.springframework.web.servlet.mvc.ParameterizableViewController"> <property name = "viewName" value="user"/> </bean> </beans>
<%@ page contentType="text/html; charset=UTF-8" %> <html> <head> <title>Hello World</title> </head> <body> <h2>Hello World</h2> </body> </html>
完成创建源文件和配置文件后，导出应用程序.右键单击您的应用程序，使用 Export → WAR文件选项并将 TestWeb.war 文件保存在Tomcat的webapps文件夹中.
现在，启动Tomcat服务器并确保能够使用标准浏览器从webapps文件夹访问其他网页.现在，尝试一个URL - http://localhost:8080/TestWeb/index.htm ，如果Spring Web Application的一切正常，您将看到以下屏幕.
