I tried this but I ran into a problem. The output of the DirectoryGenerator only gives the filename, not the full path. When the stylesheet, which is in another directory attempts to open the file, it cannot because no XML files in the directory where the stylesheet lives. Any suggestions?
> 1. use the DirectoryGenerator as startpoint > 2. use XSLT to transform the resulting XML to your > wanted output, e.g. using > document() to get the content of the specific tag > > Regards, > > Joerg > > > I would like to be able to have cocoon take an > entire > > directory of xml files, go through it, grab the > > contents of some tag (a that all the xml documents > in > > this directory have) and output a page with the > > following information for each document in that > > directory: > > > > 1. The filename (linked to the file) > > 2. The contents of the tag from the file > > > > > > How would I do this? __________________________________________________ Do You Yahoo!? Sign up for SBC Yahoo! Dial - First Month Free http://sbc.yahoo.com --------------------------------------------------------------------- Please check that your question has not already been answered in the FAQ before posting. <http://xml.apache.org/cocoon/faq/index.html> To unsubscribe, e-mail: <[EMAIL PROTECTED]> For additional commands, e-mail: <[EMAIL PROTECTED]>
