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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