问题描述
以前培训的时候讲到过这个，整合正常的做法是导入struts2-spring-plugin.jar包，然后在struts.xml里面配置action标签的class属性时直接写创建action类的bean id的名字,而不是直接写类的路径，当然new action类是通过spring来创建bean的(不管是注解方式还是xml方式)，但是实验结果是就算不导入struts2-spring-plugin.jar包,struts.xml里面配置action标签的class属性时直接写类的路径，并且也不用spring来new action类也是可以将struts2和spring整合起来的，下面我贴下一个实验代码：(没有导入struts2-spring-plugin.jar)applicationContext.xml:<?xml version="1.0" encoding="UTF-8"?><beans xmlns="http://www.springframework.org/schema/beans" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:context="http://www.springframework.org/schema/context" xmlns:aop="http://www.springframework.org/schema/aop" 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 http://www.springframework.org/schema/aop http://www.springframework.org/schema/aop/spring-aop-2.5.xsd"> <!-- 设置需要进行Spring注解扫描的类包 --> <context:component-scan base-package="com.niit" /></beans> struts2.xml:<?xml version="1.0" encoding="UTF-8"?><!DOCTYPE struts PUBLIC "-//Apache Software Foundation//DTD Struts Configuration 2.0//EN" "http://struts.apache.org/dtds/struts-2.0.dtd"><struts><constant name="struts.devMode" value="true" /> <constant name="struts.enable.DynamicMethodInvocation" value="true" /> <package name="root" namespace="/" extends="struts-default"> <action name="user" class="com.niit.test.UserAction" ><result name="res">/res.jsp</result></action></package></struts>web.xml:<?xml version="1.0" encoding="UTF-8"?><web-app version="2.5" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"> <welcome-file-list> <welcome-file>index.jsp</welcome-file> </welcome-file-list> <filter> <filter-name>struts2</filter-name> <filter-class>org.apache.struts2.dispatcher.ng.filter.StrutsPrepareAndExecuteFilter</filter-class> </filter> <filter-mapping> <filter-name>struts2</filter-name> <url-pattern>/*</url-pattern> </filter-mapping> <!-- 定义Spring配置文件位置 --><context-param><param-name>contextConfigLocation</param-name><param-value>classpath:applicationContext.xml</param-value></context-param><!-- 对Spring容器进行实例化 --><listener><listener-class>org.springframework.web.context.ContextLoaderListener</listener-class></listener></web-app>index.jsp:<%@ page language="java" import="java.util.*" pageEncoding="UTF-8"%><%String path = request.getContextPath();String basePath = request.getScheme()+"://"+request.getServerName()+":"+request.getServerPort()+path+"/";%><!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"><html> <head> <base href="<%=basePath%>"> <title>My JSP 'index.jsp' starting page</title><meta http-equiv="pragma" content="no-cache"><meta http-equiv="cache-control" content="no-cache"><meta http-equiv="expires" content="0"> <meta http-equiv="keywords" content="keyword1,keyword2,keyword3"><meta http-equiv="description" content="This is my page"><!--<link rel="stylesheet" type="text/css" href="styles.css">--> </head> <body> <a href="user!sss.action">111</a> <br> </body></html>res.jsp:<%@ page language="java" import="java.util.*" pageEncoding="UTF-8"%><%String path = request.getContextPath();String basePath = request.getScheme()+"://"+request.getServerName()+":"+request.getServerPort()+path+"/";%><!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"><html> <head> <base href="<%=basePath%>"> <title>My JSP 'res.jsp' starting page</title> <meta http-equiv="pragma" content="no-cache"><meta http-equiv="cache-control" content="no-cache"><meta http-equiv="expires" content="0"> <meta http-equiv="keywords" content="keyword1,keyword2,keyword3"><meta http-equiv="description" content="This is my page"><!--<link rel="stylesheet" type="text/css" href="styles.css">--> </head> <body> res <br> </body></html>UserAction.java:package com.niit.test;import javax.annotation.Resource;public class UserAction {@Resourceprivate UserServiceInterface userService; public String sss(){ userService.asd(); return "res"; }public UserServiceInterface getUserService() {return userService;}public void setUserService(UserServiceInterface userService) {this.userService = userService;} }UserService.java:package com.niit.test;import org.springframework.stereotype.Service;@Servicepublic class UserService implements UserServiceInterface{ public void asd(){ System.out.println("asd"); }}UserServiceInterface.java:package com.niit.test;public interface UserServiceInterface { public void asd(); }我的疑问就是应该是struts2自己new了一个UserAction实例，但是这个实例并没有交给spring管理啊，这个UserAction实例里面的userService属性应该是不会被spring注入实例的，但是为什么UserAction里面还是拿到了userService并且没有报nullpoint错误，实验的结果说明了struts2创建的UserAction实例确实交给spring了，这是为什么呢
解决方案
不导入struts2-spring-plugin.jar包 这个是必须导入的 不导入肯定注入不了 如果你是用eclipse测试的 建议多clean几次 然后再试你在acton的某个方法输出下userService 一定为null<action name="user" class="com.niit.test.UserAction" > 首先struts2 new一个，然后交给struts2-spring-plugin 注入依赖
解决方案二：
userService不是有@Resource注解么，猜想Spring扫到这个注解了，然后也会注入其实现，具体该看看Spring如何实现component-scan的源代码吧