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- To: <email@example.com>
- Subject: Re: [xml-dev] Automated schema migration
- From: "Renaud Waldura" <firstname.lastname@example.org>
- Date: Wed, 9 Nov 2005 16:06:39 -0800
- References: <20051109220531.E7D835AC3@hottub.local.mindstep.com>
Michael and others:
Thank you for your answers. I understand that it is not be possible to infer
relationships between 2 schemas, but I was referring to a transformation
Let me re-express my initial message, because I feel it might have been
understood. (I also have a thick skull, so I apologize if I'm the one who's
not understanding you.)
Say I have a schema for the old document set, and a transformation that
creates the new schema from the old -- a transformation between schemas.
This transformation creates the new schema from the old.
Could I somehow derive a transformation between documents from this
transformation between schemas?
PS: I didn't know about mapping tools, I'll look into them.
----- Original Message -----
From: "Michael Kay" <email@example.com>
To: "'Renaud Waldura'" <firstname.lastname@example.org>; <email@example.com>
Sent: Wednesday, November 09, 2005 2:01 PM
Subject: RE: [xml-dev] Automated schema migration
>> My question is: assuming I've created a schema for my old set
>> of documents,
>> if I were to create a transformation T1 that, given the old schema,
>> generates the new schema, could I machine-generate from that
>> T1 another transformation T2 that could be applied to the documents
> It's easy to demonstrate that this is impossible in general. Given a
> <p>Some <i>text</i> in italics</p>
> and another schema for
> <para>Some <italic>text</italic> in italics</para>
> It's clearly impossible to deduce the transformation by looking at the
> schemas alone.
> There are plenty of tools out there (generally called "mappers") that
> you to design a transformation starting from two schemas, by showing the
> relationships between the elements in each. They only work where the two
> schemas are quite similar, but that's probably true in your case.
> Michael Kay
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