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Re: A Query about View Construction in XML

• To: jenglish@f..., xml-dev@l...
• Subject: Re: A Query about View Construction in XML
• From: "Wayne Steele" <xmlmaster@h...>
• Date: Fri, 12 Apr 2002 10:13:55 -0700
• Bcc:

This has in fact been implemented, just as the original poster described, by 
my former employer (http://www.nimble.com).
The query language was not XSLT or XQuery, but a version of XML-QL.

-Wayne Steele


>From: Joe English <jenglish@f...>
>To: xml-dev@l...
>Subject: Re:  A Query about View Construction in XML
>Date: Thu, 11 Apr 2002 07:49:40 -0700
>
>Dong Hui Zhuo wrote:
> >
> > Given a XML database DB, and a query Q1, either XQuery or XSLT, we refer
> > to the result of Q1(DB) (the result of running query Q1 against database
> > DB) as a view, say V, whose schema is Sv.
> >
> > Suppose we have another query Q2 (either XQuery or XSLT) which is
> > formulated against schema Sv (the schema of the view V) and the
> > size of view V too big to fit in memory, my questions are:
> >
> > 1) Is it possible that we evaluate Q2(V), (that is running Q2 against 
>the
> > result of Q1(DB)), without materializing (or write back to hard disk) 
>the
> > result of Q1(DB)?
> > 2) Does the answer depend on the type of query language, say XQuery or
> > XSLT?
>
>It depends on the query language and the implementation.
>XQuery certainly has the potential to implement this efficiently.
>The query language is closed under composition, so Q2(Q1(DB)) can be
>expressed as (Q2 . Q1)(DB).  Depends on how well the implementation
>can optimize (Q2 . Q1).
>
>For XSLT I think the answer is "only for a very limited set of
>queries", and determining if Q1 and Q2 are members of that set
>is so difficult that it's unlikely to be implemented by any
>XSLT processors.
>
>
>--Joe English
>
>   jenglish@f...
>


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