If there's no template in your stylesheet that matches a given element, then
the default template kicks in: this outputs the text of the element (without
any tags). Change the default by including
<xsl:template match="*"/>
Michael Kay
http://www.saxonica.com/
-----Original Message-----
From: Ramona Krickan [mailto:ramona(_dot_)krickan(_at_)zooplus(_dot_)de]
Sent: 08 March 2007 14:52
To: 'xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com'
Subject: [xsl] XSLT Output contains the unwanted content of
all elements as text
Hi,
I got a Problem with my transformation output. My xslt doc
does the correct transformation but in my output document I
not only get the expected transformation but also the
text-content of all the other elements from my source
document. I don't undestand why. Could somebody please help me?
Here my xslt:
<xsl:stylesheet version="2.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
xmlns:zoo="http://example.com/ns";>
<xsl:template match="element(zoo:country)"> <xsl:variable
name="countryCode" select="."/> <xsl:value-of select="if
($countryCode='CH') then ('AT') else ($countryCode)" />
</xsl:template> </xsl:stylesheet>
My input xml:
<zoo:address xmlns:zoo="http://example.com/ns";>
<zoo:street>Strasse</zoo:street>
<zoo:country>CH</zoo:country>
</zoo:address>
output document:
Strasse /*this is unexpected*/
AT
Thanks in advance for any help!
Ramona Krickan
zooplus AG
Eichenweg 4a
85774 Unterfoehring
Germany
Tel +49.89.95 00 6-194
Fax +49.89.95 00 6-500
mailto:ramona(_dot_)krickan(_at_)zooplus(_dot_)de
http://www.zooplus.de
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