15 Nov
2010
15 Nov
'10
5:44 p.m.
Hello fellow developers,
in order for the current source file to be a real source file outside of the XML of the
xar, e.g. a .properties, a .vm, a .groovy, it would be enough to enable XInclude on the
parser that reads the XML document.
Which source should I look into?
Also, where for xwiki 1.5?
thanks in advance
paul
15 Nov
15 Nov
6:12 p.m.
There is a tool in the sandbox which can take any xml file and unpack it into a directory
tree.
If run on an xwiki xml file from the .xar, this tree will have a file in it called content
which
will contain all of the content of the document. Not sure if that's what you are
looking for but
it's there.
http://svn.xwiki.org/svnroot/xwiki/contrib/sandbox/xwiki-dirtree/
Caleb
On 11/15/2010 11:44 AM, Paul Libbrecht wrote:
Hello fellow developers,
in order for the current source file to be a real source file outside of the XML of the
xar, e.g. a .properties, a .vm, a .groovy, it would be enough to enable XInclude on the
parser that reads the XML document.
Which source should I look into?
Also, where for xwiki 1.5?
thanks in advance
paul
_______________________________________________
devs mailing list
devs(a)xwiki.org
http://lists.xwiki.org/mailman/listinfo/devs
6:14 p.m.
On Mon, Nov 15, 2010 at 17:44, Paul Libbrecht <paul(a)activemath.org> wrote:
Hello fellow developers,
in order for the current source file to be a real source file outside of the XML of the
xar, e.g. a .properties, a .vm, a .groovy, it would be enough to enable XInclude on the
parser that reads the XML document.
Which source should I look into?
Also, where for xwiki 1.5?
thanks in advance
You can find the maven xar plugin on
http://svn.xwiki.org/svnroot/xwiki/platform/xwiki-tools/trunk/xwiki-xar-plu…
paul
_______________________________________________
devs mailing list
devs(a)xwiki.org
http://lists.xwiki.org/mailman/listinfo/devs
--
Thomas Mortagne
5150
days inactive
5150
days old
2 comments
3 participants
participants (3)
-
Caleb James DeLisle -
Paul Libbrecht -
Thomas Mortagne