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From Jan.Mate...@rzf.fin-nrw.de
Subject AW: Getting the name of the current file in a fileset
Date Wed, 04 Jun 2003 10:30:16 GMT
I think the only way is to use AntContrib┬┤s <foreach>.
This topic is already discussed on
http://marc.theaimsgroup.com/?l=ant-user&m=104464267224692&w=2

Maybe we can add a property ${xmlfile} which is set by <xslt> and can be
used in nested <param> tags.
<xslt dir="" xsl="">
    <param name="thefile" value="${xmlfile}"/>
</xslt>
So the stylesheet can use its "thefile".

Maybe a little work in XSLTProcess.process(File baseDir, String xmlFile,
File destDir, File stylesheet). 


Jan



Ant-Contrib:
  Homepage:   http://sourceforge.net/projects/ant-contrib/
  CVS-Source:
http://cvs.sourceforge.net/cgi-bin/viewcvs.cgi/ant-contrib/ant-contrib/src/n
et/sf/antcontrib/
  CVS-Manual:
http://cvs.sourceforge.net/cgi-bin/viewcvs.cgi/ant-contrib/ant-contrib/manua
l/index.html
	
http://cvs.sourceforge.net/cgi-bin/viewcvs.cgi/*checkout*/ant-contrib/ant-co
ntrib/manual/tasks/index.html
  Binary:     http://gump.covalent.net/jars/latest/ant-contrib/



> -----Urspr├╝ngliche Nachricht-----
> Von: martin.me.roberts@bt.com [mailto:martin.me.roberts@bt.com]
> Gesendet am: Mittwoch, 4. Juni 2003 12:10
> An: user@ant.apache.org
> Betreff: Getting the name of the current file in a fileset
> 
> Hi, 
>         I need to get the current file name so I can pass it 
> into the xslt
> task as a value for the stylesheet.  How do I do this?  Is 
> there a property
> for this?
> 
> 	Martin Roberts 
> 
> 
> 

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