Subject: Re: [xsl] new bie question " opening child XML files fails" From: chun ji <cji_work@xxxxxxxxx> Date: Mon, 4 Dec 2006 18:20:02 -0800 (PST) |
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 > > ____________________________________________________________________________________ Cheap talk? Check out Yahoo! Messenger's low PC-to-Phone call rates. http://voice.yahoo.com
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