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From: Palmer, D. <Dav...@ph...> - 2003-05-08 18:55:12
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This is what we do... we use
getClass().getResourceAsStream("foo.properties");
to load our properties file from the classpath, therefor we don't have to
worry about the actual locations of said files.
Dunno if this would work in your situation, but it works out nicely for us.
I hope this helps
./dave
> -----Original Message-----
> From: Anthony Frey [mailto:ant...@co...]
> Sent: Thursday, May 08, 2003 1:21 PM
> To: wra...@li...
> Subject: [Wrapper-user] Wrapper's current working directory
>
>
>
> First off, I must say this is one of the nicest Java service
> wrappers I have
> yet to use. It's flexibility and ease-of-use are unmatched!
> Kudos and thanks
> to Leif and the Wrapper team for making it available.
>
> I think I may be missing something with configuration that
> I'm hoping someone
> has some suggestions for.
>
> My standard practice has always been to write config files
> for my Java
> applications such that paths are relative to the "home"
> directory of the
> application. I also like to put executables (including
> Wrapper) in a "bin"
> directory. I know Wrapper is designed to always use the
> executable directory
> as the working directory and that relative paths can be used
> in wrapper.conf
> which is great because I can count on it.
>
> My problem is that all of my non-wrapper.conf config files
> also need to be
> "aware" of the location of the executable and specify
> relative paths with a
> '../' prefix. The solution I've come up with is to add a
> "app.home" system
> property that gets used as the base to all relative paths
> throughout my
> application. This is a little cumbersome though since this
> needs to be done
> everywhere throughout the application. Is it possible to set the Java
> process's notion of the current working directory to this
> "home" directory or
> am I missing something else entirely?
>
> Thanks for the help,
> -Tony
>
>
>
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