Re: [xsl] Java list extension

> probeer select=*.xml
thanks, but that would miss "*.xsl".

And the question I would like to get answered is, whether it is possible 
in
XSLT 2 (or XSLT 3) to get the names of Non-XML files by the stylesheet, 
too.


Mit besten Gruessen / Best wishes,

Hermann Stamm-Wilbrandt
Developer, XML Compiler, L3
Fixpack team lead
WebSphere DataPower SOA Appliances
https://www.ibm.com/developerworks/mydeveloperworks/blogs/HermannSW/
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Vorsitzender des Aufsichtsrats: Martin Jetter
Geschaeftsfuehrung: Dirk Wittkopp
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From:   Michel Hendriksen <michel(_dot_)hendriksen(_at_)42(_dot_)nl>
To:     xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Date:   07/05/2011 02:16 PM
Subject:        Re: [xsl] Java list extension



probeer select=*.xml

On Tue, Jul 5, 2011 at 2:14 PM, Michel Hendriksen
<michel(_dot_)hendriksen(_at_)42(_dot_)nl> wrote:
> probeer select=*.xml
>
> On Tue, Jul 5, 2011 at 2:09 PM, Hermann Stamm-Wilbrandt 
<STAMMW(_at_)de(_dot_)ibm(_dot_)com> wrote:
> > Andew,
> >
> > > You can get a list of all the files in the directory and
> > > subdirectories by doing:
> > > <xsl:for-each
select="collection('file:///path/to/dir/?select=*;recurse=yes;on-error=ignore')">
> > >   <xsl:value-of select="document-uri(.)"/>
> > how can you get the names of non-XML files?
> > Saxon reports errors for applying document-uri() to Non-XML files.
> >
> > [stammw@br8ggx73 xslt20]$ ll
> > total 12
> > -rw-rw-r-- 1 stammw stammw 403 Jul  5 14:04 ls.xsl
> > -rw-rw-r-- 1 stammw stammw   5 Jul  5 14:03 some.xml
> > -rw-rw-r-- 1 stammw stammw   4 Jul  5 14:04 tst.txt
> > [stammw@br8ggx73 xslt20]$
> > [stammw@br8ggx73 xslt20]$ saxon ls.xsl some.xml
> > Error on line 1 column 1 of tst.txt:
> >  SXXP0003: Error reported by XML parser: Content is not allowed in
> > prolog.
> > file:/home/stammw/xslt/xslt20/ls.xsl
> > file:/home/stammw/xslt/xslt20/some.xml
> > [stammw@br8ggx73 xslt20]$
> > [stammw@br8ggx73 xslt20]$ cat ls.xsl
> > <xsl:stylesheet version="2.0"
> >  xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
> > >
> >  <xsl:output omit-xml-declaration="yes" />
> >
> >  <xsl:template match="/">
> >    <xsl:for-each
select="collection('file:///home/stammw/xslt/xslt20/?select=*;recurse=yes;on-error=ignore')">
> >      <xsl:value-of select="document-uri(.)"/><xsl:text>&#10;</xsl:text>
> >    </xsl:for-each>
> >  </xsl:template>
> >
> > </xsl:stylesheet>
> > [stammw@br8ggx73 xslt20]$
> >
> >
> > Mit besten Gruessen / Best wishes,
> >
> > Hermann Stamm-Wilbrandt
> > Developer, XML Compiler, L3
> > Fixpack team lead
> > WebSphere DataPower SOA Appliances
> > https://www.ibm.com/developerworks/mydeveloperworks/blogs/HermannSW/
> > ----------------------------------------------------------------------
> > IBM Deutschland Research & Development GmbH
> > Vorsitzender des Aufsichtsrats: Martin Jetter
> > Geschaeftsfuehrung: Dirk Wittkopp
> > Sitz der Gesellschaft: Boeblingen
> > Registergericht: Amtsgericht Stuttgart, HRB 243294
> >
> >
> >
> > From:   Andrew Welch <andrew(_dot_)j(_dot_)welch(_at_)gmail(_dot_)com>
> > To:     xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
> > Date:   07/05/2011 10:15 AM
> > Subject:        Re: [xsl] Java list extension
> >
> >
> >
> > > Now I need a way to discover the contents of a directory (files and
> > > other directorys). So far I have:
> > You can get a list of all the files in the directory and
> > subdirectories by doing:
> >
> > <xsl:for-each
select="collection('file:///path/to/dir/?select=*;recurse=yes;on-error=ignore')">
> >  <xsl:value-of select="document-uri(.)"/>
> >
> >
> > --
> > Andrew Welch
> > http://andrewjwelch.com
> >
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