TinkerGraph: why mutating through GraphTraversalSource doesn't persist/store while using Graph does?

Question

Hi I am trying to use TinkerGraph for some small cases of demonstration and it seems that it doesn't persist using GraphTraversalSource. Here is the code I currently have:

    GraphTraversalSource g = TinkerGraph.open().traversal();
    System.out.println(g.V().addV("testlabel").iterate());
    System.out.println(g.V().count().next().intValue()); //returns 0
    try {g.close(); }
    catch(Exception e){
        e.printStackTrace();
    }
    System.out.println(g.V().count().next().intValue()); //returns 0

and here is the output:

[TinkerGraphStep(vertex,[]), AddVertexStep({label=[testlabel]})]
0
0

I know that this works instead:

Graph graph = TinkerGraph.open();
Vertex gremlin = graph.addVertex("testlabel");
System.out.println(IteratorUtils.count(graph.vertices()) == 1);

Thank you :)

score 1 · Accepted Answer · answered Dec 29 '16 at 12:36

It looks like you need to use addV right from the GraphTraversalSource object not from V() (At least for the first object). The following code persists the vertexes in the gremlin console

gremlin> g = TinkerGraph.open().traversal();
==>graphtraversalsource[tinkergraph[vertices:0 edges:0], standard]
gremlin> g.V().addV("testing")
gremlin> g.V().count()
==>0
gremlin> g.V().addV("testing")
gremlin> g.V().count()
==>0
gremlin> g.addV("test");
==>v[0]
gremlin> g.V().count()
==>1
gremlin> g.addV("test2");
==>v[1]
gremlin> g.V().count()
==>2
gremlin> g.V().addV("testing3")
==>v[2]
==>v[3]
gremlin> g.V().count()
==>4

oh my bad! Thanks! I forgot that ```V()``` from the ```V().count()``` — Michail Michailidis, Dec 29 '16 at 12:49

TinkerGraph: why mutating through GraphTraversalSource doesn't persist/store while using Graph does?

1 Answers1