[xsl] <xsl:output doctype-system="name_of_DTD"/>

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.



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