How to remove an edge and add a new edge between two vertices?

Question

I\'m trying to drop an edge and add a new edge between two vertices. How do I do that in Tinkerpop3?

def user = g.V().has(\"userId\", \'iamuser42\').has(\"t         


        

Answer 1

The Gremlin Console tutorial discusses this issue a bit. You are not iterating your traversal. Consider the following:

gremlin> graph = TinkerFactory.createModern()
==>tinkergraph[vertices:6 edges:6]
gremlin> g = graph.traversal()
==>graphtraversalsource[tinkergraph[vertices:6 edges:6], standard]
gremlin> person = g.V().has('name','marko')
==>v[1]

Great! The person Vertex is stored in the "person" variable...or is it?

gremlin> person.class
==>class org.apache.tinkerpop.gremlin.process.traversal.dsl.graph.DefaultGraphTraversal

Apparently it is not a Vertex. "person" is a Traversal, the console sees it as such and iterates it for you which is why you get "v1" in the output. Note that the traversal is now "done":

gremlin> person.hasNext()
==>false

You need to iterate the Traversal into your variable - in this case, using next():

gremlin> person = g.V().has('name','marko').next()
==>v[1]
gremlin> person.class
==>class org.apache.tinkerpop.gremlin.tinkergraph.structure.TinkerVertex

Note that you will have further problems down the line in your script because you are using what will now be a Vertex as a Traversal, so this line:

user.outE("is_invited_to_join").where(otherV().has("groupId", 'test123')).drop();

will now fail with a similar error because Vertex will not have outE(). You would need to do something like:

g.V(user).outE("is_invited_to_join").where(otherV().has("groupId", 'test123')).drop();

If you would like to add an edge and drop the old one in the same traversal, you could do something like this:

gremlin> graph = TinkerFactory.createModern()
==>tinkergraph[vertices:6 edges:6]
gremlin> g = graph.traversal()
==>graphtraversalsource[tinkergraph[vertices:6 edges:6], standard]
gremlin> g.V(1).outE('knows')
==>e[7][1-knows->2]
==>e[8][1-knows->4]
gremlin> g.V(1).as('a').outE().as('edge').inV().hasId(2).as('b').addE('knew').from('a').to('b').select('edge').drop()
gremlin> g.V(1).outE('knew')
==>e[12][1-knew->2]
gremlin> g.V(1).outE('knows')
==>e[8][1-knows->4]

Arguably, that might not be as readable as breaking it into two or more separate traversals, but it can be done.