Query virtual graphs with graphql?

Support
1

Hi,

I'm testing the stardog7 beta on ubuntu and am still failrly new to stardog.
I have figured out how to setup a virtual graph and how to query it from studio and via http.
I have managed to get a graphql query to work on a native graph (the documentation here Home | Stardog Documentation Latest seems to be missing the @config(dataset: whatever) command.

My question:
Id like to query a virtual graph with graphql. And putting the virtual://music in the dataset parameter seems to be illegal. Could you give me a hint here?

query MyQuery @config(dataset: virtual://music) {
ALBUM {
     pp_NAME
      }
}

thank you!

2

Hi Niko,

The @config parameter you need to specify is graph: "virtual://music". You can read more in the documentation.

Best,
Jess

1 Like
3

Hi jess,

thank you for your reply! Using graph: and quotes solved the issue that it wouldn't execute. One step further!
Unfortunately, I still don't get any data, even though the "Show Plan" button describes a valid sparql query that actually returns my desired data.

My GraphQL:

query MyQuery @config(graph: "virtual://music") {
ALBUM {
        name
    }
}

My minimal virtual Graph description:

PREFIX : <http://api.stardog.com>
PREFIX owl: <http://www.w3.org/2002/07/owl#>
PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#>
PREFIX stardog: <tag:stardog:api:>
PREFIX xsd: <http://www.w3.org/2001/XMLSchema#>

MAPPING
FROM SQL {
  SELECT *
  FROM "DATA"."ALBUM"
  }
TO {
  ?subject :artist ?ARTIST .
  ?subject :name ?NAME .
  ?subject a :ALBUM
} WHERE {
  BIND(template(":/ALBUM/{ID}") AS ?subject)
}

I don't have an explicit GraphQL schema set up for the virtual graph.
Is there anything else I have to enable or to adjust to make this work?
GraphQL calls come back but with "data": []

Thank you again!

EDIT:
I did make it work finally by explicitly giving the prefixes used with the @prefix directive. From the documentation I expected that prefixes that are present in the graph wouldn't have to be declared? Any idea why that may be?

4

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