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RE: Is this valid?

From: Dare Obasanjo <dareo@microsoft.com>
Date: Mon, 22 Jul 2002 11:35:02 -0700
Message-ID: <8BD7226E07DDFF49AF5EF4030ACE0B7E06621E1E@red-msg-06.redmond.corp.microsoft.com>
To: "Jeni Tennison" <jeni@jenitennison.com>
Cc: <xmlschema-dev@w3.org>

For some reason I thought there was ambiguity over whether the minOccurs of b1 was 0 since that of its parent sequence is 0 but I realize that is incorrect. :) 

	-----Original Message----- 
	From: Jeni Tennison [mailto:jeni@jenitennison.com] 
	Sent: Mon 7/22/2002 11:12 AM 
	To: Dare Obasanjo 
	Cc: xmlschema-dev@w3.org 
	Subject: Re: Is this valid?
	
	

	Hi Dare,
	
	> I thought that but would like to know if the REC allows it given the
	> rules in
	>
	> http://www.w3.org/TR/xmlschema-1/#rcase-RecurseAsIfGroup
	> and
	> http://www.w3.org/TR/xmlschema-1/#rcase-Recurse
	>
	> Specifically can
	>
	>   <xs:sequence minOccurs="0">
	>                     <xs:element name="a1"/>
	>                     <xs:element name="b1"/>
	>    </xs:sequence>
	>
	> be restricted to
	>
	>  <xs:element name="b1"/>
	>
	> given the wording [not intentions] of the REC?
	
	When you do a restriction from a sequence to an element particle, you
	treat the element particle as if it were a sequence, so you then have
	to look at the "Recurse" rule for mapping sequences to sequences:
	
	  For an all or sequence group particle to be a ·valid restriction· of
	  another group particle with the same {compositor} all of the
	  following must be true:
	
	  1 R's occurrence range is a valid restriction of B's occurrence
	    range as defined by Occurrence Range OK (§3.9.6).
	
	  2 There is a complete ·order-preserving· functional mapping from
	    the particles in the {particles} of R to the particles in the
	    {particles} of B such that all of the following must be true:
	
	    2.1 Each particle in the {particles} of R is a ·valid
	        restriction· of the particle in the {particles} of B it maps
	        to as defined by Particle Valid (Restriction) (§3.9.6).
	
	    2.2 All particles in the {particles} of B which are not mapped
	        to by any particle in the {particles} of R are ·emptiable· as
	        defined by Particle Emptiable (§3.9.6).
	
	The crux is #2. You're asking whether, given:
	
	  (a1, b1)?  ->  (b1)
	
	there's a complete order-preserving functional mapping from a1 and b1
	to b1. The answer is that there isn't. There are various possible
	mappings, all of which fall foul of #2. The best map is:
	
	  a1  ->  nothing
	  b1  ->  b1
	
	and #2.2 says that if a particle in the base doesn't map on to
	anything (as a1 doesn't) then it must be emptiable. Emptiable means:
	
	  [Definition:] For a particle to be emptiable one of the following
	  must be true:
	  1 Its {min occurs} is 0.
	  2 Its {term} is a group and the minimum part of the effective
	    total range of that group, as defined by Effective Total Range
	    (all and sequence) (§3.8.6) (if the group is all or sequence) or
	    Effective Total Range (choice) (§3.8.6) (if it is choice), is 0.
	
	                 http://www.w3.org/TR/xmlschema-1/#cos-group-emptiable
	
	and neither of those cases is true (the min occurs for the particle a1
	is 1, and it isn't a group, it's an element particle).
	
	Do you read it differently?
	
	Cheers,
	
	Jeni
	
	---
	Jeni Tennison
	http://www.jenitennison.com/
	
	
Received on Monday, 22 July 2002 14:35:35 UTC

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