spring2.5和struts1.3.8整合

第一步:导入对应jar文件

 

第二步：

1、在web容器中实例化spring容器

<!-- 指定spring的配置文件，默认从web根目录寻找配置文件，我们可以通过spring提供的classpath：前缀指定从类路径下寻找 -->
  <context-param>
      <param-name>contextConfigLocation</param-name>
      <param-value>classpath:beans.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>

我们查看框架源码发现，spring容器被存放到ServletContext当中。

 

第三步：

在web.xml中配置struts

<servlet>
　　<servlet-name>struts</servlet-name>
　　<servlet-class>org.apache.struts.action.ActionServlet</servlet-class>
　　<init-param>
　　　　<param-name>config</param-name>
　　　　<param-value>/WEB-INF/struts-config.xml</param-value>
　　</init-param>
　　<load-on-startup>0</load-on-startup>
</servlet>
<servlet-mapping>
　　<servlet-name>struts</servlet-name>
　　<url-pattern>*.do</url-pattern>
</servlet-mapping>

 

书写struts-config.xml:

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE struts-config PUBLIC
    "-//Apache Software Foundation//DTD Struts Configuration 1.3//EN"
    "http://struts.apache.org/dtds/struts-config_1_3.dtd">
<struts-config>
    <action-mappings>
    <!--  
        <action path="/person/list" type="cn.itcast.web.action.PersonAction" validate="false">
            <forward name="list" path="/WEB-INF/page/personlist.jsp"/>
        </action>
    -->    
        <action path="/person/list" validate="false">
            <forward name="list" path="/WEB-INF/page/personlist.jsp"/>
        </action>
    </action-mappings>
    
    <!-- 加入spring请求控制器 -->
    <controller>
        <set-property property="processorClass" value="org.springframework.web.struts.DelegatingRequestProcessor"/>
    </controller>
    
</struts-config>

 

如果action没有交给spring管理时，我们通过下面语句获取spring容器实例

WebApplicationContext ctx=WebApplicationContextUtils.getWebApplicationContext(
this.getServlet().getServletContext());

 

例子代码：

 @Override
    public ActionForward execute(ActionMapping mapping, ActionForm form,
            ServletRequest request, ServletResponse response) throws Exception {
        WebApplicationContext ctx=WebApplicationContextUtils.getWebApplicationContext(
                this.getServlet().getServletContext());
        PersonService personService=(PersonService) ctx.getBean("personService");
        request.setAttribute("persons", personService.getPersons());
        System.out.println("12");
        return mapping.findForward("list");
    }

 

我们也可以将action教给spring管理：

要求：
　　1、确保action的path属性值与bean的名称相同。

　　

<action path="/person/list" validate="false">
...
@Controller("/person/list")
public class PersonAction extends Action {
    @Resource(name="personService")
    private PersonService personService;
...

 

　　2、在struts配置文件中添加进spring的请求控制器
　　该请求控制器会先根据action的path属性值到spring容器中寻找跟该属性同名的bean。如果寻找到即使用该bean处理用户请求。

<!-- 加入spring请求控制器 -->
<controller>
<set-property property="processorClass" value="org.springframework.web.struts.DelegatingRequestProcessor"/>
</controller>

 

 

较完整代码：

package cn.itcast.web.action;

import javax.annotation.Resource;
import javax.servlet.ServletRequest;
import javax.servlet.ServletResponse;

import org.apache.struts.action.Action;
import org.apache.struts.action.ActionForm;
import org.apache.struts.action.ActionForward;
import org.apache.struts.action.ActionMapping;
import org.springframework.stereotype.Controller;
import org.springframework.web.context.WebApplicationContext;
import org.springframework.web.context.support.WebApplicationContextUtils;

import cn.itcast.service.PersonService;
@Controller("/person/list")
public class PersonAction extends Action {
    @Resource(name="personService")
    private PersonService personService;
    
    

    @Override
    public ActionForward execute(ActionMapping mapping, ActionForm form,
            ServletRequest request, ServletResponse response) throws Exception {
        
        
        request.setAttribute("persons", personService.getPersons());
        System.out.println("12");
        return mapping.findForward("list");
    }
    
    
    /*
     @Override
    public ActionForward execute(ActionMapping mapping, ActionForm form,
            ServletRequest request, ServletResponse response) throws Exception {
        WebApplicationContext ctx=WebApplicationContextUtils.getWebApplicationContext(
                this.getServlet().getServletContext());
        PersonService personService=(PersonService) ctx.getBean("personService");
        request.setAttribute("persons", personService.getPersons());
        System.out.println("12");
        return mapping.findForward("list");
    }
     */

}

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE struts-config PUBLIC
    "-//Apache Software Foundation//DTD Struts Configuration 1.3//EN"
    "http://struts.apache.org/dtds/struts-config_1_3.dtd">
<struts-config>
    <action-mappings>
    <!--  
        <action path="/person/list" type="cn.itcast.web.action.PersonAction" validate="false">
            <forward name="list" path="/WEB-INF/page/personlist.jsp"/>
        </action>
    -->    
        <action path="/person/list" validate="false">
            <forward name="list" path="/WEB-INF/page/personlist.jsp"/>
        </action>
    </action-mappings>
    
    <!-- 加入spring请求控制器 -->
    <controller>
        <set-property property="processorClass" value="org.springframework.web.struts.DelegatingRequestProcessor"/>
    </controller>
    
</struts-config>

<?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">
  <display-name></display-name>    
  
  <welcome-file-list>
    <welcome-file>index.jsp</welcome-file>
  </welcome-file-list>
  
  <!-- 指定spring的配置文件，默认从web根目录寻找配置文件，我们可以通过spring提供的classpath：前缀指定从类路径下寻找 -->
  <context-param>
      <param-name>contextConfigLocation</param-name>
      <param-value>classpath:beans.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>
  
  <servlet>
      <servlet-name>struts</servlet-name>
      <servlet-class>org.apache.struts.action.ActionServlet</servlet-class>
      <init-param>
          <param-name>config</param-name>
          <param-value>/WEB-INF/struts-config.xml</param-value>
      </init-param>
      <load-on-startup>0</load-on-startup>
  </servlet>
  <servlet-mapping>
      <servlet-name>struts</servlet-name>
      <url-pattern>*.do</url-pattern>
  </servlet-mapping>
  
</web-app>

<%@ page language="java" import="java.util.*" pageEncoding="UTF-8"%>
<%@ taglib prefix="c" uri="http://java.sun.com/jsp/jstl/core" %>
<%
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>
    
    
    <title>人员列表</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>
    <c:forEach items="persons" var="person">
        ID=${person.id },name=${person.name}
    </c:forEach>
  </body>
</html>

 

 

其实我的代码没有执行成功!我不太会struts1，所以。。。仅供参考。