spring中不能找到applicationContext.xml,郁闷阿

Question

AppContext.java:package com.laoer.bbscs.sys;import org.springframework.context.support.*;/**
 * <p>Title: TianYi BBS</p>
 *
 * <p>Description: TianYi BBS System</p>
 *
 * <p>Copyright: Copyright (c) 2004</p>
 *
 * <p>Company: LAOER.COM/TIANYISOFT.NET</p>
 *
 * @author laoer
 * @version 6.0
 */
public class AppContext {  private static AppContext instance;  private AbstractApplicationContext appContext;  public synchronized static AppContext getInstance() {
    if (instance == null) {
      instance = new AppContext();
    }
    return instance;
  }  private AppContext() {
    this.appContext = new ClassPathXmlApplicationContext(
        "/applicationContext.xml");
  }  public AbstractApplicationContext getAppContext() {
    return appContext;
  }
}

Answer 1

楼上错误真实代码如下（不过差不多的哈，偶是照抄的  ：））：package com.uubbs.sys;import org.springframework.context.support.*;
public class AppContext {  private static AppContext instance;  private AbstractApplicationContext appContext;  public synchronized static AppContext getInstance() {
    if (instance == null) {
      instance = new AppContext();
    }
    return instance;
  }  private AppContext() {
    this.appContext = new ClassPathXmlApplicationContext("/applicationContext.xml");
  }  public AbstractApplicationContext getAppContext() {
    return appContext;
  }
}

Answer 2

错误如下：javax.servlet.ServletException: IOException parsing XML document from class path resource [applicationContext.xml]; nested exception is java.io.FileNotFoundException: Could not open class path resource [applicationContext.xml]
org.apache.struts.action.RequestProcessor.processException(RequestProcessor.java:545)
org.apache.struts.action.RequestProcessor.processActionPerform(RequestProcessor.java:486)
org.apache.struts.action.RequestProcessor.process(RequestProcessor.java:274)
org.apache.struts.action.ActionServlet.process(ActionServlet.java:1482)
org.apache.struts.action.ActionServlet.doPost(ActionServlet.java:525)
javax.servlet.http.HttpServlet.service(HttpServlet.java:709)
javax.servlet.http.HttpServlet.service(HttpServlet.java:802)

Answer 3

this.appContext = new ClassPathXmlApplicationContext(
        "/applicationContext.xml");使用绝对路径D:\\Eclipse\\workspace\\OA-server\\web\\WEB-INF\\applicationContext.xml

Answer 4

好像web.xml里面出现了Servlet的nested，我修改部分如下：
<context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/applicationContext.xml</param-value>
</context-param>
   <servlet>
      <servlet-name>action</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>
      <init-param>
         <param-name>debug</param-name>
         <param-value>3</param-value>
      </init-param>
      <init-param>
         <param-name>detail</param-name>
         <param-value>3</param-value>
      </init-param>
      <load-on-startup>0</load-on-startup>
   </servlet>
 <servlet>
        <servlet-name>SpringContextServlet</servlet-name>
        <servlet-class>org.springframework.web.context.ContextLoaderServlet</servlet-class>  
        <load-on-startup>1</load-on-startup>
 </servlet>