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📜  Spring MVC-参数方法名称解析器示例

📅  最后修改于: 2020-11-11 06:31:14             🧑  作者: Mango

以下示例显示了如何通过Spring Web MVC框架使用Multi Action Controller的参数方法名称解析器。 MultiActionController类有助于分别在单个控制器中映射多个URL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;
   }
}

例如，使用以上配置，如果URI-

  • /user/*.htm?action=home被请求，DispatcherServlet会将请求转发到UserController home()方法。

  • /user/*.htm?action=add被请求，DispatcherServlet会将请求转发到UserController的add()方法。

  • /user/*.htm?action=remove，DispatcherServlet会将请求转发到UserController的remove()方法。

首先，让我们拥有一个可用的Eclipse IDE，并遵循以下步骤来使用Spring Web Framework开发基于动态表单的Web应用程序。

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 a view file 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("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

user.jsp

Hello World
   
   
      
${message}

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“导出”→“ 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，并确保您能够使用标准浏览器从webapps文件夹访问其他网页。现在，尝试使用URL- http：// localhost：8080 / TestWeb / user / test.htm?action = home ，如果Spring Web Application一切正常，我们将看到以下屏幕。

弹簧多动作控制器