getting a graph path using SPARQL [duplicate]

Answer 1

There's no way to do that with sparql property paths: you just find out whether there is a path, not what the path is.

You could try:

o:a t:p{*} ?middle .
?middle t:p{*} o:o .

but I'm not convinced that will work.

Edit: ok, that does work (I thought * would be greedy), although you won't get the properties. I use !ex:nothing* to indicated 'any property' (there must be a better way):

prefix ex: <http://invalid.org/>
select * {
  o:a !ex:nothing* ?middle .
  ?middle !ex:nothing* o:o
}

Answer 2

This will give u path between a & n nodes. select ?start ?end (count(?mid) as ?length) where {values (?start ?end) { (:a :n) } ?start :p+ ?mid . ?mid :p* ?end . }group by ?start ?end