I tried that, based on Dennis' proposition:
URL url = this.getClass().getClassLoader().getResource( "/foo" );
try {
File servlet = new File(url.toURI());
getLog().info(servlet.getAbsolutePath());
} catch (URISyntaxException e) {
throw new MojoExecutionException(e.getMessage(),e);
}
But then I got an exception on the first line of the try block:
java.lang.IllegalArgumentException: URI is not hierarchical
I was thinking of using IOUtils to copy the content of foo to another
directory but it seems to be harder than I thought.
2006/7/16, dan tran <[EMAIL PROTECTED]>:
Thanks
On 7/16/06, Dennis Lundberg <[EMAIL PROTECTED]> wrote:
>
> It depends on what you want to do with the resource.
>
> I used it with
> FileUtils.copyURLToFile( url, new File( outputDirectory, filename ) );
> to copy a resource from within the jar file to the target directory.
>
> --
> Dennis Lundberg
>
> dan tran wrote:
> > Dennis, would you suggestion work? since the resource is in a jar,
> > and therefore obtaining a "File" object is not possible.
> >
> > am I missing something?
> >
> > -D
> >
> >
> > On 7/16/06, Dennis Lundberg <[EMAIL PROTECTED]> wrote:
> >>
> >> Sebastien Arbogast wrote:
> >> > In the plugin I'm working on, I have a src/main/resources/foo
> >> > directory and as a result, when the plugin is packaged, I get a foo
> >> > directory at the root of the plugin jar, and that's fine.
> >> >
> >> > But now I want to get a java.io.File reference to this foo directory
> >> > from inside the code of one of my mojos. How can I do that?
> >> >
> >>
> >> You can get a URL which can then be used to create a file like this:
> >>
> >> URL url = this.getClass().getClassLoader().getResource( "/foo" );
> >>
> >>
> >> --
> >> Dennis Lundberg
> >>
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> >>
> >>
> >
>
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>
--
Sébastien Arbogast
http://www.sebastien-arbogast.com
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