Springmvc 简明教程

Spring MVC - Parameter Method Name Resolver Example

以下示例展示了如何使用 Spring Web MVC 框架的 Parameter Method Name Resolver 使用多动作控制器。 MultiActionController 类有助于为单个控制器分别映射多个网址及其方法。

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("user");
      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 = "com.tutorialspoint.UserController">
   <property name = "methodNameResolver">
      <bean class = "org.springframework.web.servlet.mvc.multiaction.ParameterMethodNameResolver">
         <property name = "paramName" value = "action"/>
      </bean>
   </property>
</bean>

例如，使用以上配置，如果请求 URI −

  1. /user/ .htm?action=home is requested, DispatcherServlet will forward the request to the UserController *home() 方法。

  2. /user/ .htm?action=add is requested, DispatcherServlet will forward the request to the UserController *add() 方法。

  3. /user/ .htm?action=remove is requested, DispatcherServlet will forward the request to the UserController *remove() 方法。

首先，我们准备一个可用的 Eclipse IDE，然后遵循以下步骤使用 Spring Web 框架开发基于动态表单的 Web 应用程序。

Step

Description

1

如 Spring MVC 的 Hello World 章节中所述，在 com.tutorialspoint 包下创建一个以 TestWeb 命名的项目。

2

在 com.tutorialspoint 包下创建 Java 类 UserController。

3

在 jsp 子文件夹下创建视图文件 user.jsp。

4

最后一步是创建源文件和配置文件的内容，然后按如下所述导出应用程序。

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("user");
      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.mvc.support.ControllerClassNameHandlerMapping">
      <property name = "caseSensitive" value = "true" />
   </bean>
   <bean class = "com.tutorialspoint.UserController">
      <property name = "methodNameResolver">
         <bean class = "org.springframework.web.servlet.mvc.multiaction.ParameterMethodNameResolver">
            <property name = "paramName" value = "action"/>
         </bean>
      </property>
   </bean>
</beans>

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。

现在，启动 Tomcat 服务器，并确保可以使用标准浏览器访问 webapps 文件夹中的其他网页。现在尝试一个网址 http://localhost:8080/TestWeb/user/test.htm?action=home ，如果 Spring Web 应用程序一切正常，我们将看到以下屏幕。

spring multiaction controller2