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From "Vassilis Koutavas" <>
Subject Re: Generating an aggregated xml from all the files in a directory
Date Sat, 03 Mar 2007 21:06:25 GMT
Thanks Vangeli,

So let me get this straight. Can I write:

<dir:directory xmlns:dir=""
    name="the/dir/" sort="name" reverse="false">
  <dir:file name="*.xml">
    <dir:xpath query="*"/>

to get all the contents of all files in the/directory? (I'm sorry but
I'm ignorant about xpath)

Then the question is how do I put this in a forrest's aggregation. Or
I don't need an aggregation if I use this?


On 3/3/07, Evangelos Vlachogiannis <> wrote:
> Hi Vasili,
> I think you need Xpathdirectory Generator
> ( I
> hope this is included in forrest..
> Then you just have a root element (elements) and call an xpath
> expression to include what ever elements you need from xmls in dir.
> hope that helps,
> Vangelis
> Vassilis Koutavas wrote:
> > What is the right way to create one aggregated xml file from all the
> > xml files in a directory?
> > I think I need to use map:aggregate, map:part, and dir:directory in my
> > sitemap.xml, but I don't quite understand how to put them together. I
> > don't know how many files will be in the directory, thus I don't know
> > how many parts the aggregation will have.
> >
> > Let's say that in my xdocs I have the directory called "the/dir/" in
> > which there are a number of xml files, and I want to create a file
> > called "the-list.xml" (in the same direcory) which will contain the
> > contents of all the xml files in that directory, under a common
> > element called "elements". After I have this I know how to transform
> > it using an xsl, and produce the final html file.
> >
> > Best
> --
> ---------------------------------------------------------------
> Evangelos Vlachogiannis
> Researcher - PhD. Candidate
> Contact:
> ---------------------------------------------------------------


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