Springmvc 简明教程

Spring MVC - Multi Action Controller Example

以下示例展示了如何使用Spring Web MVC框架使用Multi Action Controller。 MultiActionController 类帮助分别将多个URL与其方法映射到单个控制器。

The following example shows how to use the Multi Action Controller using the Spring Web MVC framework. The MultiActionController class helps to map multiple URLs with their methods in a single controller respectively.

package com.tutorialspoint;

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("home");
      model.addObject("message", "Home");
      return model;
   }

   public ModelAndView add(HttpServletRequest request,
      HttpServletResponse response) throws Exception {
      ModelAndView model = new ModelAndView("user");
      model.addObject("message", "Add");
      return model;
   }

   public ModelAndView remove(HttpServletRequest request,
      HttpServletResponse response) throws Exception {
      ModelAndView model = new ModelAndView("user");
      model.addObject("message", "Remove");
      return model;
   }
}
<bean class = "org.springframework.web.servlet.handler.BeanNameUrlHandlerMapping"/>
<bean name = "/home.htm" class = "com.tutorialspoint.UserController" />
<bean name = "/user/*.htm" class = "com.tutorialspoint.UserController" />

例如，使用以上配置，如果请求 URI −

For example, using the above configuration, if URI −

  1. /home.htm is requested, DispatcherServlet will forward the request to the UserController home() method.

  2. user/add.htm is requested, DispatcherServlet will forward the request to the UserController add() method.

  3. user/remove.htm is requested, DispatcherServlet will forward the request to the UserController remove() method.

首先，让我们准备一个可用的Eclipse IDE，并坚持以下步骤，使用Spring Web框架开发基于动态表单的Web应用程序。

To begin with, let us have a working Eclipse IDE in place and stick to the following steps to develop a Dynamic Form based Web Application using the Spring Web Framework.

Step

Description

1

Create a project with a name TestWeb under a package com.tutorialspoint as explained in the Spring MVC - Hello World chapter.

2

Create a Java class UserController under the com.tutorialspoint package.

3

Create view files home.jsp and user.jsp under the jsp sub-folder.

4

The final step is to create the content of the source and configuration files and export the application as explained below.

UserController.java

package com.tutorialspoint;

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("home");
      model.addObject("message", "Home");
      return model;
   }

   public ModelAndView add(HttpServletRequest request,
      HttpServletResponse response) throws Exception {
      ModelAndView model = new ModelAndView("user");
      model.addObject("message", "Add");
      return model;
   }

   public ModelAndView remove(HttpServletRequest request,
      HttpServletResponse response) throws Exception {
      ModelAndView model = new ModelAndView("user");
      model.addObject("message", "Remove");
      return model;
   }
}

TestWeb-servlet.xml

<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.BeanNameUrlHandlerMapping"/>
   <bean name = "/home.htm"
      class = "com.tutorialspoint.UserController" />
   <bean name = "/user/*.htm"
      class = "com.tutorialspoint.UserController" />
</beans>

home.jsp

<%@ page contentType = "text/html; charset = UTF-8" %>
<html>
   <head>
      <meta http-equiv = "Content-Type" content = "text/html; charset = ISO-8859-1">
      <title>Home</title>
   </head>

   <body>
      <a href = "user/add.htm" >Add</a> <br>
      <a href = "user/remove.htm" >Remove</a>
   </body>
</html>

user.jsp

<%@ page contentType = "text/html; charset = UTF-8" %>
<html>
   <head>
      <title>Hello World</title>
   </head>
   <body>
      <h2>${message}</h2>
   </body>
</html>

一旦你创建完资源文件和配置文件，导出你的应用程序。右键点击你的应用程序，使用 Export → WAR File 选项并在Tomcat的webapps文件夹中保存 TestWeb.war 文件。

Once you are done with creating source and configuration files, export your application. Right click on your application, use Export → WAR File option and save the TestWeb.war file in Tomcat’s webapps folder.

现在，启动您的Tomcat服务器，并确保您可以使用标准浏览器从webapps文件夹访问其他网页。现在，尝试URL − http://localhost:8080/TestWeb/home.htm ，如果Spring Web应用程序一切正常，我们将看到以下屏幕。

Now, start your Tomcat server and make sure you are able to access other webpages from webapps folder using a standard browser. Now, try a URL − http://localhost:8080/TestWeb/home.htm and we will see the following screen, if everything is fine with the Spring Web Application.

spring multiaction controller

尝试URL http://localhost:8080/TestWeb/user/add.htm ，如果Spring Web应用程序一切正常，我们将看到以下屏幕。

Try a URL http://localhost:8080/TestWeb/user/add.htm and we will see the following screen, if everything is fine with the Spring Web Application.

spring multiaction controller1