Re: [xsl] accessing xsl document root from an included stylesheet

[Jan Bilik <jan.bilik.w@xxxxxxxxx>]

2009/12/4 Michael Kay <mike@xxxxxxxxxxxx>:
> It's both verbose and inefficient compared with the much simpler
>
> <xsl:variable name="xslmyn" select="document('')/*/namespace::myn" />

Thanks ! I'll fix this.

> (..) But there's no way of disovering in an
> included module where it was included from.

I read on w3c.org that "the resource located by the href attribute
value is parsed as an XML document, and the children of the
xsl:stylesheet element in this document replace the xsl:include
element in the including document."
Thus I thought the included code could access namespaces aliases of
the main xsl:stylesheet element. Anyway, I was clearly wrong.

> However, it all seesm very long-winded to me. Why don't you just do
>
> <xsl:template match="/">
>  <xsl:if test="not(myn:rootElement)">
>     <xsl:message terminate="yes">Root element must be named myn:rootElement
> in the correct namespace</xsl:message>

I think I can not use this alternative, as my grammar is not the main
one : it defines additional attributes and elements that expand the
base grammars (to be clear : XSD and WADL are the base grammars).
Thus, I have no idea of what is the root element, which grammar it
fits, etc. In fact, I know it for *each* of my XSLs, of course, but
not in my case, as I was looking for a generic template to check the
namespace version.

Thanks a lot for your detailed answer and advice, Michael.

Regards,

Jan

>
> Regards,
>
> Michael Kay
> http://www.saxonica.com/
> http://twitter.com/michaelhkay