Re: [xsl] XPath and noname namespace

Subject: Re: [xsl] XPath and noname namespace
From: George Cristian Bina <george@xxxxxxxxxxxxx>
Date: Mon, 14 Jul 2008 18:08:14 +0300
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="";>
than you will get different results.

Best Regards,
George Cristian Bina
<oXygen/> XML Editor, Schema Editor and XSLT Editor/Debugger

Adam Komisarek wrote:

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="";>
<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


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