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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