Will it not work without an entry in web.xml?
Is it mandatory in tomcat 5 version?
Shud i make an entry for each servlet that i use to submit in my html?

Vj

----- Original Message -----
From: "Chris Pratt" <chrisp@(protected)>
To: <SERVLET-INTEREST@(protected)>
Sent: Monday, March 29, 2004 9:48 AM
Subject: Re: requested resource (/quiz/servlet/quiz) is not available


> Try going to http://majid:8080/quiz
>
>  (*Chris*)
>
> -----Original Message-----
> From: A mailing list for discussion about Sun Microsystem's Java Servlet
API
> Technology. [mailto:SERVLET-INTEREST@(protected)
Nakit
> Sent: Sunday, March 28, 2004 7:43 pm
> To: SERVLET-INTEREST@(protected)
> Subject: [SERVLET-INTEREST] requested resource (/quiz/servlet/quiz) is not
> available
>
> I have a servlet called Quiz.java , I put it and compile it in :
> C:\jakarta-tomcat-5.0.19\webapps\quiz\WEB-INF\classes
> and added web.xml in :
> C:\jakarta-tomcat-5.0.19\webapps\quiz\WEB-INF\
>
> In web.xml I have :
>
> <?xml version="1.0" encoding="UTF-8"?>
> <!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>Quiz</servlet-name>
>   <servlet-class>Quiz</servlet-class>
>  </servlet>
>  <servlet-mapping>
>   <servlet-name>Quiz</servlet-name>
>   <url-pattern>/quiz</url-pattern>
>  </servlet-mapping>
> </web-app>
>
> BUT when I started tomcat and put in the server :
> http://majid:8080/quiz/servlet/quiz
>
> I got this error :
>
> HTTP Status 404 - /quiz/servlet/quiz
>
> type Status report
>
> message /quiz/servlet/quiz
>
> description The requested resource (/quiz/servlet/quiz) is not available.
> Apache Tomcat/5.0.19
>
> Please help me.
> thanks
>
>
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