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From Dietmar.Muel...@eurotours.at
Subject Re: [users@httpd] Tomcat 3.2.3 vs Tomcat 5.5.9 issue
Date Thu, 29 Sep 2005 10:40:12 GMT

This is the wrong list!

But why do you use backslash within the path from the form?
Try slash instead.

regards Dietmar





Cyndi Rader <crader@Mines.EDU> am 29.09.2005 00:15:13

Bitte antworten an users@httpd.apache.org

An:    users@httpd.apache.org
Kopie:
Thema: [users@httpd] Tomcat 3.2.3 vs Tomcat 5.5.9 issue


Hi,

I recently downloaded Tomcat 5.5.9 on a Windows XP machine and
moved a small servlet application from 3.2.3 to it.  This is
actually an example from a textbook.  Unfortunately the servlet
will not run under 5.5.9.  The web.xml file entries for this
servlet are:

  <servlet>
    <servlet-name>welcome1</servlet-name>
      <description>
      A simple servlet that handles an HTTP get request.
     </description>
    <servlet-class>
      com.deitel.advjhtp1.servlets.WelcomeServlet
    </servlet-class>
  </servlet>
  <servlet-mapping>
    <servlet-name>welcome1</servlet-name>
    <url-pattern>/welcome1</url-pattern>
  </servlet-mapping>

The html form that calls this includes the following:
   <form action = "\advjhtp1\welcome1" method = "get">

My directory structure includes:
<inst-dir>
  <webapps>
    <advjhtp1>
       <WEB-INF>
          web.xml
          <classes>
            <com deitel etc.>
               WelcomeServlet.class

The message I get when I try to run is Status 404, the requested
resource is not available.  As I said, I copied the advjhtp1
directory from webapps under a 3.2.3 Tomcat installation, where it
ran fine.  So I'm wondering what the difference might be?  FYI, I
did have to recompile the class, because it first gave the message
com.deitel.advjhtp1.servlets.WelcomeServlet is not a Servlet. So I
recompiled using the servlet-api.jar file under Tomcat5.5\common\lib,
which resolved the "not a Servlet" issue, but left the status 404
code. The servlet itself basically just prints "Welcome to Servlets".
Any suggestions would be greatly appreciated!


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