Subject: [xsl] XSLT1.0 xml-stylesheet into XML output From: "Kirov Plamen" <pkirov@xxxxxxxxx> Date: Fri, 20 Oct 2006 14:26:53 +0300 |
Hello, I'm having the following: XSL: <?xml version="1.0" encoding="UTF-8"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output method="xml" encoding="UTF-8" doctype-public="-//W3C//DTD HTML 4.01 Transitional//EN" indent="no"/> <!-- root --> <xsl:template match="/"> <!-- Inv Part --> <Inv> <Acc> <Name>ABCDEF</Name> <FamName>XYZ</FamName> </Acc> </Inv> </xsl:template> </xsl:stylesheet> Generated by this XSL XML: <?xml version="1.0" encoding="UTF-8"?> <Inv> <Acc> <Name>ABCDEF</Name> <FamName>XYZ</FamName> </Acc> </Inv> But I want to have the following XML as output: <?xml version="1.0" encoding="UTF-8"?> <?xml-stylesheet type="text/xml" href="Name.xsl"?> <Inv> <Acc> <Name>ABCDEF</Name> <FamName>XYZ</FamName> </Acc> </Inv> What changes into XSL is needed? Regards, Plamen
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