On Fri, 20 Jun 2003, Xu, John wrote:

> Milt,
> Thanks for your response.
> The purpose to do this is to find the current directory to open a
> property file which is located in a directory that this program is
> running.

You may want to do a search on this (either on the web, newsgroups, or
this list's archives), because this is a pretty common thing to do.
"property file" or "configuration file" would be good search terms.

But I think you want to use getResource(), or getResourceAsStream() --
that abstracts the file to a "resource", and you don't have to worry
about filesystem directories and paths.  Jason Hunter discusses it in
the following article (it's a chapter from the book Java Enterprise
Best Practices):

http://www.onjava.com/pub/a/onjava/excerpt/jebp_3/index1.html

Also, remember, "this program" is Tomcat, so the "directory that this
program is running" is where Tomcat was started from.  And that's
probably not what you want.


> -----Original Message-----
> From: Milt Epstein [mailto:[EMAIL PROTECTED]
> Sent: Friday, June 20, 2003 5:00 PM
> To: [EMAIL PROTECTED]
> Subject: Re: Find current directory of java program running
>
>
> On Fri, 20 Jun 2003, Xu, John wrote:
>
> > Hi,
> > For a jsp page or servlet running within Tomcat, what is the good
> > way to find the directory that it is running?  Can we use
> > System.getProperty("user.dir")?
>
> For the best answer, you should describe why you want to do this
> and/or what are you trying to do.  Then we can tell you the
> best/cleanest way to do it.
>
> Milt Epstein
> Research Programmer
> Integration and Software Engineering (ISE)
> Campus Information Technologies and Educational Services (CITES)
> University of Illinois at Urbana-Champaign (UIUC)
> [EMAIL PROTECTED]
>
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Milt Epstein
Research Programmer
Integration and Software Engineering (ISE)
Campus Information Technologies and Educational Services (CITES)
University of Illinois at Urbana-Champaign (UIUC)
[EMAIL PROTECTED]

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