Reading Gerrit's note I realize that in my answer I was mixing XSLT
with XSD in the declaration of the element.
His way of using prefix-from-QName() is correct, mine is not. I
should have taken the time to test it, then I would have realized I
was answering the wrong question.
. . . . . . . . Ken
At 2012-08-27 16:41 -0400, I wrote:
At 2012-08-27 20:37 +0000, Costello, Roger L. wrote:
Hi Folks,
I want to transform this:
<xs:attribute name="test" />
to this:
<xs:attribute name="test" type="xs:anySimpleType" />
Here is the code that I tried:
<xsl:template match="xs:attribute">
<xsl:copy>
<xsl:copy-of select="@*" />
<xsl:attribute name="type"><xsl:value-of
select="QName('http://www.w3.org/2001/XMLSchema',
'anySimpleType')" /></xsl:attribute>
<xsl:sequence select="node()" />
</xsl:copy>
</xsl:template>
Unfortunately that code produces this:
<xs:attribute name="test" type="anySimpleType" />
Notice that there is no prefix on anySimpleType.
What's the best way to accomplish my objective?
I want the prefix on anySimpleType to match the prefix on the
<attribute> element, whether it be xs: or xsd: or anything else.
Retrieve it from the namespace axis:
(untested; assumes the default namespace is not the XSD namespace)
<xsl:attribute
name="{name(namespace::*[.='http://www.w3.org/2001/XMLSchema'])]}:anySimpleType"
namespace="http://www.w3.org/2001/XMLSchema">
<xsl:copy-of select="node()"/>
</xsl:attribute>
or in XSLT 2.0:
(untested)
<xsl:attribute name="{prefix-from-QName(node-name(.))}:anySimpleType"
namespace="http://www.w3.org/2001/XMLSchema">
<xsl:copy-of select="node()"/>
</xsl:attribute>
Remember that node-name(.) gives the QName of the name of the
current node, not the name string of the name of the current node.
I hope this helps.
. . . . . . . . . Ken
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