Gremlin get Edge with specific start and end node

Question

I am using Gremlin to handle Titan Graph. And I am trying to find a way to get a very specific relationship.

I have the label, the properties and a list of possible start and endNodes.

And I want all relationships matching this.

I have this already to get all relationships matching label and property:

GraphTraversal<Edge, Edge> tempOutput = g.E().hasLabel(relationshipStorage.getId());

            if(relationshipStorage.getProperties() != null)
            {
                for (Map.Entry<String, Object> entry : relationshipStorage.getProperties().entrySet())
                {
                    if (tempOutput == null)
                    {
                        break;
                    }
                    tempOutput = tempOutput.has(entry.getKey(), entry.getValue());
                }
            }

But I didn't find a way to get it with a specific start and endNode. I don't want to get multiple edges between two nodes. I only want one edge with the specific vertices.

stephen mallette stephen mallette · Accepted Answer · 2016-10-31T10:46:21

See the Between Vertices recipe and just extend from there. For example, let's say you wanted to find edges between two vertices with ids 1 and 2. Let's further assume you only wanted "knows" edges with a "weight" property greater than 0.0.

gremlin> graph = TinkerFactory.createModern()
==>tinkergraph[vertices:6 edges:6]
gremlin> g = graph.traversal()
==>graphtraversalsource[tinkergraph[vertices:6 edges:6], standard]
gremlin> g.V(1).bothE().where(otherV().hasId(2)).hasLabel('knows').has('weight',gt(0.0))
==>e[7][1-knows->2]
gremlin> g.V(1,2).bothE().where(inV().has(id, within(2,3))).hasLabel('created')
==>e[9][1-created->3]
gremlin> vStarts = g.V(1,2).toList().toArray()
==>v[1]
==>v[2]
gremlin> vEnds = g.V(2,3).toList().toArray()
==>v[2]
==>v[3]
gremlin> g.V(vStarts).bothE().where(inV().is(within(vEnds))).hasLabel('created')
==>e[9][1-created->3]

Gremlin get Edge with specific start and end node

1 Answers