TinkerPop3 Gremlin遍历

Question

First, Given a Node A, first we need to find out all nodes that could be reached through the "Friend" edges(A,B,C... all the blue nodes). I can achieve this by using the path()-step like bellow:

 g.V().hasLabel("Person").has("name", "A")
                            .repeat(out("Friend"))
                            .until(out("Friend").count().is(0))
                            .path();

And then we can extract verticals from the path objects in java code.

But actually, we need to find out what books they read(the green ones). But we can't extract the verticals from path() in gremlin.

Is there any way we could do this in one gremlin traversal?

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Edit: In fact, there is 2 situations:

1. from A, find out all persons connected through "Friend" edge. We did this by the traversal mentioned above. (is there any better way? ie. extract these nodes directly in gremlin?)

2. from A, find out all persons then all books they read, and return the books, only the books.

Answer 1

You could use the union step.

g.V()
 .hasLabel("person")
 .has("name", "A")
 .repeat(union(out("friend"),
               out("read")))
 .until(out("friend").count().is(0))
 .union(path(),
        out("read").path())

Should get you what you need. With a graph defined as

gremlin> :> graph.addVertex(T.label, "person", "name", "A")
==>v[0]
gremlin> :> graph.addVertex(T.label, "person", "name", "B")
==>v[2]
gremlin> :> graph.addVertex(T.label, "person", "name", "C")
==>v[4]
gremlin> :> graph.addVertex(T.label, "person", "name", "D")
==>v[6]
gremlin> :> a = g.V(0).next(); b = g.V(2).next(); a.addEdge("friend", b, "is", "is");
==>e[8][0-friend->2]
gremlin> :> b = g.V(2).next(); c = g.V(4).next(); a.addEdge("friend", b, "is", "is");
==>e[9][2-friend->4]
gremlin> :> a = g.V(0).next(); d = g.V(6).next(); a.addEdge("friend", b, "is", "is");
==>e[10][0-friend->6]
gremlin> :> graph.addVertex(T.label, "book", "name", "Huck Finn")
==>v[11]
gremlin> :> graph.addVertex(T.label, "book", "name", "Tom Sawyer")
==>v[13]
gremlin> :> graph.addVertex(T.label, "book", "name", "A Tale Of Two Cities")
==>v[15]
gremlin> :> a = g.V(0).next(); b = g.V(11).next(); a.addEdge("read", b, "is", "is");
==>e[17][0-read->11]
gremlin> :> a = g.V(2).next(); b = g.V(13).next(); a.addEdge("read", b, "is", "is");
==>e[18][2-read->13]
gremlin> :> a = g.V(4).next(); b = g.V(15).next(); a.addEdge("read", b, "is", "is");
==>e[19][4-read->15]

It yields

gremlin> :> g.V().hasLabel("person").has("name", "A").repeat(union(out("friend"), out("read"))).until(out("friend").count().is(0)).union(path(), out("read").path())
==>[v[0], v[6]]
==>[v[0], v[11]]
==>[v[0], v[2], v[4]]
==>[v[0], v[2], v[4], v[15]]
==>[v[0], v[2], v[13]]