Ok, it is in fact working, the namespace doesn't appear with
exclude-result-prefixes="xs". My mistake was that i still have my
original problem, so that means that it's not a XSL problem, so i'm
back to the drawing board...
Thanks for your help.
On 10/18/05, Michael Kay <mike(_at_)saxonica(_dot_)com> wrote:
The rules for a literal result element say that all its in-scope namespaces
are copied to the result tree.
However, you should be able to prevent this using exclude-result-prefixes.
If this isn't working, please supply a complete (but small!) stylesheet to
demonstrate the problem so we can see what you are doing wrong.
Michael Kay
http://www.saxonica.com/
-----Original Message-----
From: António Mota [mailto:amsmota(_at_)gmail(_dot_)com]
Sent: 18 October 2005 00:59
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] Namespace on output node but not not in source
Hi:
I have a xsl with a namespace declared
<xsl:transform xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
version="1.0"
xmlns:xs="http://www.w3.org/2001/XMLSchema";>
so i can make some lookups on a xsd file
<xsl:variable name="tipo"
select="$schema/xs:schema/xs:element[(_at_)name=$nome]/@type"/>
and that's all i do, i don't copy any node from $schema to
the output tree.
However, this
<xsl:template match="/">
<table>
<xsl:apply-templates
select="(//Menu)[position()=$pos]"/>
</table>
</xsl:template>
produces a
<table xmlns:xs="http://www.w3.org/2001/XMLSchema";>
<...>
</table>
and i don't understand why, and i don't want it there.
I add a exclude-result-prefixes="xs" but it seems it makes no
diference.
Why is this?
Thanks.
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