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Message-ID: <029f01c14c2e$79c11c70$0301a8c0@white>
From: "Bob Byron" <bbyron@radit.com>
To: <tomcat-user@jakarta.apache.org>, <craigmcc@apache.org>
References: <20011003092700.P49410-100000@localhost>
Subject: Re: Open a file?
Date: Wed, 3 Oct 2001 12:11:38 -0500
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Thank you Craig!  

Of course I just sent out a renewed plea for help.  Murphy's
Law.  I checked for email, sent out my new plea, and 
then received your email.

I will be utilizing this technique.  Thanks for the info.

Bob Byron

----- Original Message ----- 
From: "Craig R. McClanahan" <craigmcc@apache.org>
To: <tomcat-user@jakarta.apache.org>
Sent: Wednesday, October 03, 2001 11:28 AM
Subject: Re: Open a file?


> 
> 
> On Tue, 2 Oct 2001, Bob Byron wrote:
> 
> > Date: Tue, 2 Oct 2001 22:03:37 -0500
> > From: Bob Byron <bbyron@radit.com>
> > Reply-To: tomcat-user@jakarta.apache.org
> > To: tomcat-user@jakarta.apache.org
> > Subject: Open a file?
> >
> > Okay, now I need to open the XML file and I am trying to
> > figure out how to do that.  When I try to open the file "file.xml",
> > Tomcat 4 is trying to open it relative to "Tomcat\4.0\bin", the
> > location of the startup scripts.  I want to know what is the
> > Tomcat centric way of opening a file relative to the webapp
> > that it is running under?  I hope I have phrased the question
> > sufficiently for you to understand what I need to do.  I need
> > a different reference point, one that uses the webapp itself
> > as a base.
> >
> 
> If you have a file "file.xml" in the top-level directory of your web app,
> open it like this:
> 
>   InputStream stream =
> getServletContext().getResourceAsStream("/file.xml");
> 
> This will work on all servlet containers (not just Tomcat).  It will also
> work when your web app is run directly from a WAR file rather than being
> unpacked.
> 
> > Thank You,
> > Bob Byron
> >
> >
> 
> Craig McClanahan
>