Subject: Re: XSLT1.0 xml-stylesheet into XML output
From: "Andrew Welch" <andrew.j.welch@xxxxxxxxx>
Date: Fri, 20 Oct 2006 12:47:23 +0100
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On 10/20/06, Kirov Plamen <pkirov@xxxxxxxxx> wrote:
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?
Have a read of:
http://www.dpawson.co.uk/xsl/sect2/N6145.html#d8258e76
cheers
andrew
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