Re: [xsl] namespace-uri() method returns emty,using xalan 2.0

[Oleg Tkachenko <olegt at multiconn dot com>]

challa sumalatha wrote:

> Hey list:
> i tried with all of the following options to get the namespace-uri(),all of my efforts are in vain.In all of these cases it returns empty.Help needed....

> xml:
> <?xml version="1.0"?>
>  <a:root xmlns:a="name-a">
>  	<b:sub xmlns:b="name-b"/>
>  	<c:sub xmlns:c="name-c"/>
>  </a:root>
> xsl:method1:
> <?xml version="1.0"?>
>  <xsl:stylesheet
>  xmlns:xsl="http://www.w3.org/1999/XSL/Transform"; 
>  version="1.0"
>    xmlns:a="name-a"  xmlns:b="name-b"  
>  xmlns:c="name-c">
>  <xsl:template match="/">
>    <xsl:for-each select="a:root">
>     <uri><xsl:value-of select="namespace-uri()"/></uri>
>   </xsl:for-each>
>  </xsl:template>
>  </xsl:stylesheet>

This one works fine for me on msxml3 and saxon and because of its 
simplicity it have to work :) I got
<uri xmlns:a="name-a" xmlns:b="name-b" xmlns:c="name-c">name-a</uri>


> xsl: method:2
> <?xml version="1.0"?>
>  <xsl:stylesheet
>  xmlns:xsl="http://www.w3.org/1999/XSL/Transform"; 
>  version="1.0"
>    xmlns:a="name-a"  xmlns:b="name-b"  
>  xmlns:c="name-c">
>  <xsl:template match="/">
>    <xsl:for-each select="*/[name()='a:root']">

This is XPath syntax error - predicate always requires nodetest, you 
probably mean *[name()='a:root'], which is tough version of a:root.
I guess the problem is in your processor, which one are you using?


-- 
Oleg Tkachenko
Multiconn International, Israel


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