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Subject: [xsl] get the name of a node which I know the value From: Stephane.Le-Deaut@xxxxxxxxxx Date: Mon, 20 Aug 2001 17:41:06 +0200 |
I have this xml file :
<?xml version="1.0" encoding="ISO-8859-1"?>
<XslParameter>
<InputFile
currentAuthentifiedPackageDescriptor1="C:/XML/currentAuthentifiedPackageDescriptor_N1.xml"
currentAuthentifiedPackageDescriptor2="C:/XML/currentAuthentifiedPackageDescriptor_N2.xml"
referenceAuthentifiedPackageDescriptor_Nectar="C:/XML/referenceAuthentifiedPackageDescriptor_Nectar.xml"
automate="C:/XML/automate.xml"
user="C:/XML/user_prg_dha.xml"/>
<OutputFile
subsetDescriptor="C:/XML/subsetDescriptor.xml"
xslTrack="C:/XML/XslTrack_Test_makeSubsetDescriptor_01.xml"
xslError="C:/XML/XslError_Test_makeSubsetDescriptor_01.xml"/>
<Parameter
signature="C:/XML/Signature.xml"/>
</XslParameter>
I know the value of the "currentAuthentifiedPackageDescriptor2" attribute
and I would like to assign a variable with the name of the node.
I tried to use name() function like this
<xsl:variable name="CurrentFile"
select="name()[starts-with(@*,'C:/XML/currentAuthentifiedPackageDescriptor_N2.xml')]"/>
but it does not work.
Please help me.
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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