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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
>
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