You write
<xsl:template match="element(*, my_type)">
and use a schema-aware XSLT 2.0 processor.
Michael Kay
http://www.saxonica.com/
-----Original Message-----
From: David B [mailto:daavidb(_at_)gmail(_dot_)com]
Sent: 17 July 2006 16:57
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] Matching elements based on element type:
Hello,
I have an xml file that looks like this:
<top>
<a>123</a>
<b>456</b>
<c>789</c>
</top>
And has a schema as follows:
<?xml version="1.0" encoding="UTF-8"?>
<xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema";>
<xs:element name="top">
<xs:complexType>
<xs:sequence>
<xs:element name="a" type="my_type"/>
<xs:element name="b" type="my_type"/>
<xs:element name="c" type="my_type"/>
</xs:sequence>
</xs:complexType>
</xs:element>
<xs:simpleType name="my_type">
<xs:restriction base="xs:string"/>
</xs:simpleType>
</xs:schema>
That is, the elements a,b,c all have the type my_type but
have different element names.
I want an xslt stylesheet that has a single template for all
elements of type my_type. Something like:
<xsl:template match="top">
<xsl:apply-templates/>
</xsl:template>
<xsl:template match="*[type()=my_type]">
<xsl:value-of select="."/>
</xsl:template>
Only there is no type() function in XPath....
Is there any other way I could do this?
Best Regards,
David Belius
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