[dsdl-discuss] Re: Namespace transformation

At 2006-09-14 19:43 +0100, Martin Bryan wrote:
>The problem seems to be that attributes in the xmlns namespace can't be
>picked up using @*. How can I get them to transfer indirectly?

In XSLT you address namespaces by looking along the namespace:: axis.

   <xsl:for-each select="namespace::*">
      Prefix: <xsl:value-of select="name(.)"/>
      URI: <xsl:value-of select="."/>
   </xsl:for-each>

If I have a variable named "x" that has a URI, I find the namespace node by:

    select="namespace::*[.=$x]"

If I have a variable named "y" that has a prefix I find the namespace node by:

    select="namespace::*[name(.)=$y]"

Here is some logic (from the
http://www.oasis-open.org/committees/document.php?document_id=16765
package) where I synthesize Schematron <ns>
declarations that will handle all namespaces for
any document, without relying on prefix-based
processing (except for grabbing the code list
namespace from the stylesheet), just URI-based processing:

<!--this shortcut makes some namespace work more succinct in the stylesheet-->
<xsl:variable name="clc-namespace" select="document('')/*/namespace::clc"/>

<!--work from the associated files-->
<xsl:variable name="namespaces" select="$assoc-file//namespace::*"/>

<!--declare namespaces assumed to be used for XPath addresses-->
<xsl:for-each select="$namespaces">
   <xsl:if test="generate-id(.)=
                 generate-id($namespaces[name(.)=name(current())][1]) and
                 name(.) != 'xml' and . != $clc-namespace" >
     <ns uri="{.}" prefix="{name(.)}"/>
     <xsl:copy-of select="$indent"/>
   </xsl:if>
</xsl:for-each>

A lot can be done with XSLT 1.0 and the namespace
axis ... for example, the above code in
Schematron stylesheets might have prevented the
need to create the "<ns>" element in the language (but it is there now).

I haven't looked at the rest of the message, this
just jumped out at me as something quick to
answer as that is all I can manage these days.

I hope this helps.

. . . . . . . . . Ken

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