On Wed, 2006-11-22 at 11:23 +1000, Cameron McCormack wrote:
> Hi again.
>
> I’m calling forrest with an exec from my ant build.xml, but I’d like
> Forrest’s project.build-dir property to be set based on some properties
> in my build file. To test it out first, I’ve been trying this from the
> command line:
>
> $ forrest -Dproject.build-dir=blah site
>
> but with that, Forrest tells me:
>
> ...
> X [0] linkmap.html BROKEN:
> /usr/local/apache-forrest-0.7/main/webapp/blah/tmp/build-info.xml (No such
> file or directory)
>
> as if it is resolving the relative path "blah" with a base of
> main/webapp from within Forrest’s installation directory. In my
> forrest.properties, I notice that the default value of project.build-dir
> is ${project.home}/build, so I guess ${project.home} is initialised to
> be the current directory when Forrest starts. I can’t access that
> ${project.home} variable from the command line though.
>
Hmm, works for me with and but normally doing:
forrest -Dproject.home=`pwd`/exporter
should work.
> Any suggestion on how I can make the build go under the current
> directory (apart from doing something like `pwd` in command line, since
> it needs to be cross platform)?
<antcall target="forrest.site">
<param name="project.home" location="${exporter.home}" />
<param name="project.start-uri"
location="/summaries/${bulletin.year}/${bulletin.number}/index.html" />
<param name="project.build-dir"
location="${build.dir}/exporter/summaries"></param>
</antcall>
I have a main ant file in my customer which is the parent of
${exporter.home}.
tree .
`--exporter
By changing the project.home I can invoke the site generation from my
base app.
salu2
>
> Thanks,
>
> Cameron
>
