Re: [xsl] XSLT1.0 xml-stylesheet into XML output

Subject: Re: [xsl] XSLT1.0 xml-stylesheet into XML output
From: "Andrew Welch" <andrew.j.welch@xxxxxxxxx>
Date: Fri, 20 Oct 2006 12:47:23 +0100
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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