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Re: Servlet Questions

From: David P. White <dpwhite@halcyon.com>
Date: Mon, 14 Jun 1999 12:16:45 -0400 (EDT)
Message-ID: <37652B71.3688@halcyon.com>
To: Benoit Mahe <Benoit.Mahe@sophia.inria.fr>
CC: www-jigsaw@w3.org
Benoit Mahe wrote:
> The servletwrapper is not an alias, it's the way Jigsaw handle servlets.
> So yes, all servlets must be referenced via a ServletWrapper. But,
> you can call a servlet with is full classname.
> For example, it the servlet com.foobar.servlets.TestServlet is in your CLASSPATH
> 
> or under a ServletDirectory, you reach it at
> 
> http://hostname:port/servlet/com.foobar.servlets.TestServlet
> 
> A ServletWrapper, ("com.foobar.servlets.TestServlet"), will be
> created by the ServletIndexer.

I have tried to call a servlet using it's full classname as indicated
above. I get the following displayed in my browser:

Invalid URL

The URL
http://localhost:8001/servlet/hmssoftware.utility.JACLInterpreterPoolServlet
that you requested is not available on that server

The servlet's class file is located on the classpath passed to java
using the -cp option when jigsaw is run. It is NOT located under
jigsaw's servlet directory.

Has anyone been able to execute servlets in this fashion?

Thanks,

David
Received on Monday, 14 June 1999 12:42:12 GMT

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