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
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