Subject: [xsl] <xsl:output doctype-system="name_of_DTD"/> From: Stephane.Le-Deaut@xxxxxxxxxx Date: Mon, 14 May 2001 11:51:18 +0200 |
Hello, I've two questions : Can I specify dynamically the name of the DTD ? I've already tried a solution which consist to revover this name in an attribute of a file XML. Example of the beginning of my procedure XSL : <xsl:variable name="sgDeliveryOutFileLocation" select="//XslParameter/InputFile/@sgDeliveryOutFile"/> <!-- variable to be used to access the sgDeliveryOutFileLocation input file --> <xsl:variable name="sgDeliveryOutFile" select="document($sgDeliveryOutFileLocation)"/> <!-- definition of the output format --> <xsl:output method="xml" omit-xml-declaration="no" encoding="ISO-8859-1"/> <xsl:output doctype-system="$sgDeliveryOutFile/CriteriaTargetList/CriteriaTarget/@identity"/> <xsl:output indent="yes"/> <xsl:strip-space elements="*"/> But it doesn't work and I obtain this result : <?xml version="1.0" encoding="ISO-8859-1"?> <!DOCTYPE Delivery SYSTEM "$sgDeliveryOutFile/CriteriaTargetList/CriteriaTarget/@identity"> And I would like to have this result : <?xml version="1.0" encoding="ISO-8859-1"?> <!DOCTYPE Delivery SYSTEM "name_of_DTD (maybe elNotOcbDelivery.dtd or ocbDelivery.dtd ....)"> Is it possible to recover this name in an attribute of a file XML ? Thanks for your help. XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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