Re: [xsl] XPath and noname namespace

[George Cristian Bina <george@xxxxxxxxxxxxx>]

Because the local name is the same as the name in your example for the a 
element.
If you will have
<s:svg xmlns:s="http://www.w3.org/2000/svg";>
  <s:a>aadsf</s:a>
</s:svg>
than you will get different results.

Best Regards,
George
--
George Cristian Bina
<oXygen/> XML Editor, Schema Editor and XSLT Editor/Debugger
http://www.oxygenxml.com

Adam Komisarek wrote:
> Hello!
>
> I have found workaround but I wonder why it must be done in that way..
> Is it some kind of bug in XPath?
>
> I have simple document:<svg xmlns="http://www.w3.org/2000/svg";>
> <a>aadsf</a> </svg>
>
> And now "//a" doesn't find a node.. I must use //*[local-name() =
> 'a'], but funny thing is that //*[name() = 'a'] works as well.. Why is
> that?
>
> Thanks,
> Adam