一个简单问题。。。

报错内容:type Exception reportmessage description The server encountered an internal error () that prevented it from fulfilling this request.exception javax.servlet.ServletException: Servlet.init() for servlet action threw exception
org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:118)
org.apache.coyote.tomcat5.CoyoteAdapter.service(CoyoteAdapter.java:160)
org.apache.coyote.http11.Http11Processor.process(Http11Processor.java:799)
org.apache.coyote.http11.Http11Protocol$Http11ConnectionHandler.processConnection(Http11Protocol.java:705)
org.apache.tomcat.util.net.TcpWorkerThread.runIt(PoolTcpEndpoint.java:577)
org.apache.tomcat.util.threads.ThreadPool$ControlRunnable.run(ThreadPool.java:683)
java.lang.Thread.run(Thread.java:619)
root cause java.lang.NullPointerException
org.apache.struts.action.ActionServlet.parseModuleConfigFile(ActionServlet.java:705)
org.apache.struts.action.ActionServlet.initModuleConfig(ActionServlet.java:670)
org.apache.struts.action.ActionServlet.init(ActionServlet.java:329)
javax.servlet.GenericServlet.init(GenericServlet.java:211)
org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:118)
org.apache.coyote.tomcat5.CoyoteAdapter.service(CoyoteAdapter.java:160)
org.apache.coyote.http11.Http11Processor.process(Http11Processor.java:799)
org.apache.coyote.http11.Http11Protocol$Http11ConnectionHandler.processConnection(Http11Protocol.java:705)
org.apache.tomcat.util.net.TcpWorkerThread.runIt(PoolTcpEndpoint.java:577)
org.apache.tomcat.util.threads.ThreadPool$ControlRunnable.run(ThreadPool.java:683)
java.lang.Thread.run(Thread.java:619)
note The full stack trace of the root cause is available in the Apache Tomcat/5.0.28 logs.

解决方案 »

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java.lang.NullPointerException
空指针。
<%@ page contentType="text/html;charset=gb2312" %>
<html>
<body>
<form name="f" action="welcome.do" method="post">
姓名：<input type=text name="user"/><br>
密码：<input type=password name="pwd" />
<input type="submit" value="click"/>
</form>
</body>
</html>
WEB.XML:<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>2</param-value>
    </init-param>
    <init-param>
      <param-name>detail</param-name>
      <param-value>2</param-value>
    </init-param>
    <load-on-startup>2</load-on-startup>
  </servlet>  
  <servlet-mapping>
    <servlet-name>action</servlet-name>
    <url-pattern>*.do</url-pattern>
  </servlet-mapping>STRUTS-CONFIG.XML:
<action
             path="/welcome"
      type="LoginAction"
            validate="false">
         <forward name="fail" path="/error.jsp" />
         <forward name="success" path="/success.jsp" />
    </action>ACTION：
public ActionForward execute(ActionMapping mapping,ActionForm form,HttpServletRequest request,HttpServletResponse response)
throws IOException, ServletException{
//如果用户名为scott，密码为tiger，则登录成
String userName=request.getParameter("user");
String password=request.getParameter("pwd");
System.out.print("aaaaaaaaaaaa");
if (userName!=null && password!=null && userName.equals("manager") && password.equals("1234")) {
HttpSession session = request.getSession();
session.setAttribute("loggedIn", "1");
return mapping.findForward("success");
}else{
return mapping.findForward("fail");
}
}高手们帮我看看哈子，谢了
什麼沒傳過去啊type Exception report 這是樣式異常。發生在：javax.servlet.ServletException: Servlet.init() for servlet action threw exception這就意味著，這個servlet的init（）都沒通過。你的ActionForm form　配置在什麼地方？
上面给出的代码没有错误，我在本地执行OK。
你那里执行不通过，我想还是查一查tomcat的设置，以及web-inf/classes文件夹下编译过的类的版本。
tomcat设置问题在那？
java version 1.6.0-02
这个一般来讲是因为配置文件没找到，打出个war包，仔细检查一把
/WEB-INF/struts-config.xml