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

[Christian Sell <christian.sell@xxxxxxxxxxxxx>]

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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