<xsl:element name="foo" namespace="$x"
is your friend
david
Yes, but this would require creating all the output with <xsl:element>
The OP (in another forum) wants this default namespace to apply to
literal result elements that are descendents of this top node.
However, they are just copied to the output 1:1 and the serializer
takes special care to express the fact that they belong to no
namespace, by inserting " xmlns='' " on all of them.
It seems to me that using <xsl:namespace> it is not possible to
specify a default namespace.
I would be very glad if someone provides a concrete code sample
proving me wrong :)
Cheers,
Dimitre
On Fri, Mar 12, 2010 at 11:26 AM, David Carlisle
<davidc(_at_)nag(_dot_)co(_dot_)uk> wrote:
On 12/03/2010 19:15, Dimitre Novatchev wrote:
http://www.w3.org/TR/xslt20/#element-namespace
If the effective value of the name attribute is a zero-length string, a
namespace node is added for the default namespace.
so
<foo>
<xsl:namespace name="" select="$x"/>
sets the default namespace to the uri in the variable x.
David
David, I tried this hours before asking the question.
Saxon raises this error:
er because I got it wrong, sorry, I should have checked or known or
something.
You can add namespace nodes using xsl:namespace but you can't change the
namespace of a node that's already been created so
<foo>
<xsl;namespace name=""
doesn't work.
<xsl:element name="foo" namespace="$x"
is your friend
david
--
Cheers,
Dimitre Novatchev
---------------------------------------
Truly great madness cannot be achieved without significant intelligence.
---------------------------------------
To invent, you need a good imagination and a pile of junk
-------------------------------------
Never fight an inanimate object
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You've achieved success in your field when you don't know whether what
you're doing is work or play
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a messy bloodbath.
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