Re: [xsl] new bie question " opening child XML files fails"

[chun ji <cji_work@xxxxxxxxx>]

Hi Spencer, 
that does works, but somehow it gives me another
problem, my XSL script opens only the first child xml
file, and ignore the rests. here is my revised xsl
script.
"         
<xsl:variable name="subdir"
select="document('index.xml')/protest/package/name"/>

<xsl:for-each
select="document(concat(document('index.xml')/protest/package/name,
'.xml'))/suites/suite/test">
... 
</xsl:for-each>
" 

any comments ? 


cji 


--- Spencer Tickner <spencertickner@xxxxxxxxx> wrote:

> Try
> 
> concat(document('index.xml')/protest/package/name,
> '.xml')
> 
> Spencer
> 
> On 12/4/06, chun ji <cji_work@xxxxxxxxx> wrote:
> > I have an index.xml file, which reads as:
> > "
> > -<?xml version="1.0" encoding="ISO-8859-1" ?>
> > -<?xml-stylesheet href="rules.xsl"
> type="text/xsl"?>
> > -<protest>
> >     -<package>
> >            -<name>billing</name>
> >       -</package>
> >      -<package>
> >            -<name>timecard</name>
> >       -</package>
> > -</protest>
> > ".
> > There is a group of child xml files that
> associated
> > with each "/protest/package/name", such as
> > "billing.xml", "timecard.xml".
> >
> > Now I have soming in my XSL file that tries to
> open
> > all these CHILD xml files,
> > "
> > <xsl:for-each
> >
>
select="document(document('index.xml')/protest/package/name)">
> > 
> > </xsl:for-each>
> > "
> > which fails, because I did not give ".xml" for
> each
> > child xml.
> >
> > Does someone know the format to do it ?
> >
> >
> > thanks a lot
> >
> >
> > cji
> >
> >
> >
> >
>
____________________________________________________________________________________
> > Yahoo! Music Unlimited
> > Access over 1 million songs.
> > http://music.yahoo.com/unlimited
> 
> 




 
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