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Herb RamosSubject: How to omit namespace attribute
Author: Herb Ramos
Date: 08 Jun 2005 02:56 PM
Originally Posted: 08 Jun 2005 02:32 PM
Hi, I have a stylesheet which creates a namespace attribute for one of my elements, which I don't want. I'm not sure why it creates this, but I want it removed. See attached file. The result.vxml file has a bunch of elements with blank xmlns="" attributes and a <rule> element with a xmlns="http://www.w3.org/2001/vxml" attribute. I want these ommitted. I'm not sure what is going on. Any ideas? BTW: I tried creating a template for the <rule> element, but then the first children element aquired the undesired namespace.


UnknownTest(10).xsl
Style Sheet

UnknownSource.vxml
Source

UnknownResult.vxml
Result

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Minollo I.Subject: How to omit namespace attribute
Author: Minollo I.
Date: 09 Jun 2005 10:33 PM
You are using xsl:copy to copy some nodes to your result (like, for example, the root <vxml/>), and you are creating some other elements "by hand", with xsl:element instructions. xsl:copy (http://www.w3.org/TR/xslt#copying), as also xsl:copy-of, need to copy namespace declarations too; and <vxml/> belongs to the xmlns="http://www.w3.org/2001/vxml" in the source. That's why it's created that way in the result. When you create the other elements "by hand", those are defined in the default namespace, not in "http://www.w3.org/2001/vxml"; and Saxon (I'm assuming you are using Saxon, as this is an XSLT 2 stylesheet) uses the XML 1.1 semantics of xmlns="", which is to undefine the current xmlns definition.

So, I would say that the occurences of xmlns you are getting are expected; to avoid them, you'll need to avoid using xsl:copy and xsl:copy-of, as the whole source document lives in a redefined default namespace.

Hope this helps,
Minollo

Postnext
Christopher AtkinsSubject: How to omit namespace attribute
Author: Christopher Atkins
Date: 13 Jun 2005 11:01 AM
I'm not using copy or copy-of but matching a namespace qualified node in a template... I want to get rid of the namespace defintions in the output XML as they are not needed. How do I do this?

Postnext
Minollo I.Subject: How to omit namespace attribute
Author: Minollo I.
Date: 13 Jun 2005 11:14 AM
"Normal" namespace definitions can be removed from the output defining the "exclude-result-prefixes" attribute in the xsl:stylesheet element; like:
<xsl:stylesheet .... exclude-result-prefixes="prefix1 prefix2 prefix3">

See http://www.w3.org/TR/xslt#literal-result-element for more details.

Minollo

Postnext
Herb RamosSubject: How to omit namespace attribute
Author: Herb Ramos
Date: 14 Jun 2005 08:57 AM
Originally Posted: 14 Jun 2005 08:52 AM
Thanks for your input. I have <xsl:copy> in places where I want to copy all the attributes<br> of an element. I thought namespace was not an attribute node and hence would not be copied. What should I do instead of <xsl:copy> to eliminate xmlns="". Should I loop through all the attributes and manually recreate them using <xsl:attribute> ? Thanks again

Posttop
(Deleted User) Subject: How to omit namespace attribute
Author: (Deleted User)
Date: 14 Jun 2005 11:03 AM
xsl:copy-of on an element node will copy all the attributes including xmlns to the output. but if you do xsl:copy-of on attributes seperately(after you create the element), it should just copy the real attributes.

you can try this:

<br>
<xsl:copy-of select="br/@*"/>
</br>

You need to decide the xpath nased on current context, but the idea is
do xsl:copy-of on @* which represents all the real attributes

Hope this helps.

Song.

   
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