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

2006-10-20 04:47:50
On 10/20/06, Kirov Plamen <pkirov(_at_)globul(_dot_)bg> 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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