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Re: [xsl] problem - generating XML schema via XSLT


Subject: Re: [xsl] problem - generating XML schema via XSLT
From: Christian Sell <christian.sell@xxxxxxxxxxxxx>
Date: Fri, 03 Oct 2003 22:35:29 +0200

Michael Kay wrote:
Namespace declarations in the source document are translated by the XML
parser into namespace nodes, they are not treated by XSLT as attribute
nodes. Therefore, a namespace declaration can never contain an attribute
value template (because it isn't an attribute).

You need to produce in the result tree a namespace node whose name
(=prefix) is "" and whose string value (=namespace URI) is "test". (It's
very confusing that you chose to use the name "prefix" to refer to the
namespace URI!

In XSLT 2.0 you can do this with

<xsl:namespace name=""><xsl:value-of select="@prefix"/></xsl:namespace>

thats what I call a solution. I am switching to Saxon (7.7).



In 1.0 there is no direct way of doing this. The nearest equivalent is:


<xsl:variable name="dummy">
  <xsl:element name="e" namespace="{@prefix}"/>
</xsl:variable>

<xsl:copy select="xx:node-set($dummy)/*/namespace::*[.=@prefix]"/>

This creates a dummy element in the required namespace, and then copies
the required namespace node to the result tree.

this should have been xsl:copy-of, I suppose. In addition, the "[.=@prefix]" predicate does not match anything. I had to change it to


<xsl:variable name="ns_name" select="@prefix">
<xsl:copy select="xx:node-set($dummy)/*/namespace::*[.=$ns_name]"/>

to get it to work (strange. why?). After doing that, Saxon will indeed output the namespace node, while Xalan (which behaves the same regarding the predicate) does not.

thanks,
christian


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