Spring简单整合Struts2实现登录例子

1、先完成Struts2简单登录例子的实现，当前的目录结构如下

2、右击项目添加spring开发能力

3、导入struts2的spring支持包struts2-spring-plus-2.0.11.jar(注意应该与struts核心包版本一致)，建立path

4、修改web.xml,添加监听器，param-value中定义applicationContext配置路径

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi=""
 xmlns=""
  xmlns:web=".xsd"
   xsi:schemaLocation="
    .xsd" version="3.0">

  <filter>
    <filter-name>struts2</filter-name>
    <filter-class>
    	org.apache.struts2.dispatcher.FilterDispatcher
    </filter-class>
  </filter>
  <filter-mapping>
    <filter-name>struts2</filter-name>
    <url-pattern>/*</url-pattern>
  </filter-mapping>
  
  <!-- 	 spring配置文件位置  -->
	<context-param>
	<param-name>contextConfigLocation</param-name>
	<param-value>/WEB-INF/classes/applicationContext.xml</param-value>
	</context-param>
	<!-- 	spring监听器  -->
	<listener>
	<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
	</listener>
  
</web-app>

5、在applicationContext.xml中配置mainAction的Bean

<?xml version="1.0" encoding="UTF-8"?>
<beans
	xmlns=""
	xmlns:xsi=""
	xmlns:p=""
	xsi:schemaLocation=" .1.xsd">

	<bean id="mainAction" class="action.MainAction"></bean>
</beans>

6、修改在struts.xml中mainAction的class属性值为Bean的id值

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE struts PUBLIC
        "-//Apache Software Foundation//DTD Struts Configuration 2.0//EN"
        ".0.dtd">
	<struts>
		<package name="default" extends="struts-default">
			<!-- 用户登录action -->
			<action name="main" class="mainAction">
				<result name="success">/success.jsp</result>
				<result name="error">/error.jsp</result>
			</action>
		</package>
		<!-- <constant name="struts.i18n.encoding" value="gb2312"></constant> -->
	</struts>
 

7、启动Tomcat测试结果