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RE: Quick question regarding XPath

• To: "'Murali Mani'" <mani@C...>,<xml-dev@l...>
• Subject: RE: Quick question regarding XPath
• From: "Michael Kay" <michael.h.kay@n...>
• Date: Thu, 26 Feb 2004 10:47:26 -0000
• In-reply-to: <Pine.GSO.4.58.0402250549490.17412@p...>
• Thread-index: AcP7pnVxVO5xDJlCTvmLuFLhAoRltwAref0Q

# 
# is it possible to say in XPath
# 
# a//@b
# 
# is the above equivalent to:
# 
# (a/@b | a//*/@b)
# 

Yes. I think the easiest proof of this (using constructs that are allowed in
XPath 2.0 along the way) is

a//@b 

means

a/descentant-or-self::node()/@b

which means

a/(descendant::node | self::node())/@b

which means

a/descendant::node()/@b | a/self::node()/@b

which means

a/child::node()/descendant-or-self::node()/@b | a/@b

which means

a/descendant-or-self::node()/child::node()/@b | a/@b

which means (given that nodes other than elements have no attributes)

a//*/@b | a/@b

Of course this assumes some equivalences which I haven't tried to prove
formally, for example that "|" distributes over "/". 

Michael Kay

• References:
  • Quick question regarding XPath
    • From: Murali Mani <mani@C...>

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