Servlet class X is not a javax.servlet.Servlet解决方案

LiaoJuncai

浏览: 466971 次
性别:
来自: 北京

最近访客更多访客>>

imti189

u012361334

thousandlin

wang_eye

博主相关

博客

微博

相册

留言

关于我

文章分类

社区版块

存档分类

博客分类：

Java编程

写servlet的时候，报500错：Servlet class X is not a javax.servlet.Servlet

详细报错：

HTTP Status 500 -

--------------------------------------------------------------------------------

type Exception report
message

description The server encountered an internal error () that prevented it from fulfilling this request.

exception
javax.servlet.ServletException: Class com.servlet.LoginServlet is not a Servlet org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:102)
org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:298)
org.apache.coyote.http11.Http11Processor.process(Http11Processor.java:852)
org.apache.coyote.http11.Http11Protocol$Http11ConnectionHandler.process(Http11Protocol.java:588)
org.apache.tomcat.util.net.JIoEndpoint$Worker.run(JIoEndpoint.java:489)
java.lang.Thread.run(Thread.java:619)

root cause
java.lang.ClassCastException: com.servlet.LoginServlet cannot be cast to javax.servlet.Servlet org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:102)
org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:298)
org.apache.coyote.http11.Http11Processor.process(Http11Processor.java:852)
org.apache.coyote.http11.Http11Protocol$Http11ConnectionHandler.process(Http11Protocol.java:588)
org.apache.tomcat.util.net.JIoEndpoint$Worker.run(JIoEndpoint.java:489)
java.lang.Thread.run(Thread.java:619)

note The full stack trace of the root cause is available in the Apache Tomcat/6.0.26 logs.

原始代码：

import java.io.*;
import javax.servlet.*;
import javax.servlet.http.*;
public class LoginServlet {
public void doPost(HttpServletRequest request,HttpServletResponse response)
throws ServletException,IOException{
String username=request.getParameter("username");
String password=request.getParameter("password");
response.setContentType("text/html;charset=gb2312");
PrintWriter out=response.getWriter();
out.println("<html><body>");
/*if(username.equals("admin")&&password.equals("admin")){
out.println("login successfully!"+username);
}else{
out.println("sorry!Your username or password is uncorrect");
}*/
out.println("haha");
out.println("</body></html>");
}
public void doGet(HttpServletRequest request,HttpServletResponse response)
throws ServletException,IOException{
doPost(request,response);
}
}

解决方案说明：

写一个servlet时，类需要承继HttpServlet 。
例如：

public class
MyHttpServletRequest extends HttpServlet {  
    public void doGet(HttpServletRequestrequest, HttpServletResponse response) throws 
     } 
 public void doPost(HttpServletRequestrequest, HttpServletResponse response) throws 
            ServletException, IOException { 
        doGet(request, response); 
    } 
}

分享到：

Web app root system property already set ... | common-configuration读取xml,properties文 ...

2013-12-04 14:29
浏览 10027
评论(0)
分类:Web前端
查看更多

发表评论

您还没有登录,请您登录后再发表评论

最近访客更多访客>>

博主相关

文章分类

社区版块

存档分类

最新评论