CODEWITHSUNDEEP
gremlin
Michail Michailidis
Michail Michailidis

Reputation: 12191

TinkerGraph retrieval inconsistency when using ids as String vs Long

When I store a vertex in TinkerGrapg I see that the ids are long:

{TinkerVertex@7955} "v[304]"

when I do: 

graph.V(304).next();

it doesn't work!

when I do:

graph.V("304").next();

it doesn't work!

when I do: 

graph.V(304l).next();

or

graph.V(new Long(304)).next();

it works!

I am trying to use the same Gremlin code both against TinkerGraph and DSE Graph..The problem is that one returns long and the other one as strings

I am curious how I could make the same Gremlin work with int/long and String ids at the same time.. is this a problem of the API?

Thanks!

Upvotes: 0

Views: 204

Answers (1)

stephen mallette
stephen mallette

Reputation: 46226

TinkerGraph uses and IdManager which can coerce identifiers to different types. You can read more about that here but basically if you want to have g.V(1) and g.V(1L) both return a value you'd want to configure your TinkerGraph as follows:

gremlin> conf = new BaseConfiguration()
==>org.apache.commons.configuration.BaseConfiguration@552518c3
gremlin> conf.setProperty('gremlin.tinkergraph.vertexIdManager',"LONG")
gremlin> graph = TinkerGraph.open(conf)
==>tinkergraph[vertices:0 edges:0]
gremlin> g = graph.traversal()
==>graphtraversalsource[tinkergraph[vertices:0 edges:0], standard]
gremlin> g.addV(id, 1L)
==>v[1]
gremlin> g.addV(id, 2)
==>v[2]
gremlin> g.V(1)
==>v[1]
gremlin> g.V(1L)
==>v[1]
gremlin> g.V(2L)
==>v[2]
gremlin> g.V(2)
==>v[2]

In fact, even this works when you use an IdManager:

gremlin> g.V("2")
==>v[2]

You can specify your own IdManager implementation to TinkerGraph if you like - just supply the fully qualified classname to the configuration (i.e. instead of "LONG").

Upvotes: 4

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