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Subject: Re: [xdi] Realizations about $has statements
Hello Markus and Drummond, I would propose the following: =example $is$a +person =example+email $1 ?person@example.com? =example+email is the set of emails belonging to =example, $1 is the accessor for ?person@example.com?. It is easy to find =example+email$1 solving one by one the subsegments from left to right. And this is unambiguous, assumed that $has is left associative. What do you think? Kind Regards, Giovanni Def. Quota "Markus Sabadello" <markus.sabadello@xdi.org>: > So the point is that in a PDS you want to do this, right? > > =example > $is$a > +person > +email$1 > ?person@example.com? > =example$1 > $is$a > +home+person > +home+email > =example+email$1 > > Not really sure how well it would work in practice.. If there were more > subsegments than just =example and +email$1, how would I know where to break > them up into their "subject" and "predicate" parts in order to follow the > reference and find the literal value? > > I.e. in the above example, how do I know I have to look for subject =example > and then for predicate +email$1, rather than for subject =example+email and > then for predicate $1 ? > > Markus > ---------------------------------------------------------------- This message was sent using IMP, the Internet Messaging Program.
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