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I have a servlet file,like
follows:
import javax.servlet.*; import javax.servlet.http.*; import java.io.*; public class a1 extends HttpServlet{ public void doGet(HttpServletResponse req,HttpServletResponse resp) throws ServletException,IOException{ PrintWriter pw=resp.getWriter(); pw.println("<html><body>Hello World!</body></html>"); pw.close(); } } and web.xml is follows: <?xml version="1.0" encoding="ISO-8859-1"?> <!DOCTYPE web-app PUBLIC "-//Sun Microsystems, Inc.//DTD Web Application 2.3//EN" "http://java.sun.com/dtd/web-app_2_3.dtd"> <web-app> <servlet> <servlet-name>a1</servlet-name> <servlet-class>a1</servlet-class> </servlet> </web-app> I use Tomcat5.0,I don't want to use Tomcat5 default path ,so I put a1.class into \Tomcat 5.0\webapps\test\WEB-INF\classes,and I have added a sentence into server.xml,like follows: <Context path="test" docBase="test" debug="0"/> but when I run it in IE6,like follows: http://localhost:8080/test/servlets/a1 it raises HTTP Status 404 error,"The requested resource (/test/servlets/a1) is not available." Why? I have configure path! Please help. Thanks in advance Edward Archives: http://archives.java.sun.com/archives/servlet-interest.html Resources: http://java.sun.com/products/servlet/external-resources.html LISTSERV Help: http://www.lsoft.com/manuals/user/user.html
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