CODEWITHSUNDEEP
javagremlin
Michail Michailidis
Michail Michailidis

Reputation: 12191

Is there a way to always return a List of Maps for Gremlin match queries with select?

I have the following code:

    Vertex v = g.addV().property("valueStr", "3").next();
    Vertex v2 = g.addV().property("valueStr", "4").next();
    Vertex v3 = g.addV().property("valueStr", "5").next();
    Edge e = g.V(v.id()).as("a").V(v2.id()).as("b").addE("anEdge").from("a").to("b").property("value", "4").as("e").next();
    Edge e2 = g.V(v.id()).as("a").V(v3.id()).as("b").addE("anEdge").from("a").to("b").property("value", "5").as("e").next();

    List vertices1 = g.V().match(
            __.as("a").hasId(v.id()).outE("anEdge").inV().hasId(v2.id()).as("b"),
            __.as("a").hasId(v.id()).outE("anEdge").inV().hasId(v3.id()).as("c")).toList();
    System.out.println(vertices1);

    List vertices2 = g.V().match(
            __.as("a").hasId(v.id()).outE("anEdge").inV().hasId(v2.id()).as("b"),
            __.as("a").hasId(v.id()).outE("anEdge").inV().hasId(v3.id()).as("c")).select("a","b").toList();
    System.out.println(vertices2);

    List vertices3 = g.V().match(
            __.as("a").hasId(v.id()).outE("anEdge").inV().hasId(v2.id()).as("b"),
            __.as("a").hasId(v.id()).outE("anEdge").inV().hasId(v3.id()).as("c")).select("a").toList();
    System.out.println(vertices3);

It is basically node a connecting to b, and node a connecting to c.

I am doing a match query over this subgraph and each time I return a subset of the elements that were matched..

Here are the outputs:

[{a=v[20], b=v[22], c=v[24]}]
[{a=v[20], b=v[22]}]
[v[20]]

In the first two cases I get a List of Maps.. In the last case I get a List.

How is it possible for the last case to make it also a List of Maps or [{a=v[20}]? I know I can do a hack with select("a","a") but it seems there should be a cleaner way.

Where does the documentation explain in which cases I will be getting a List of Vertices/Edges, a Map or a List of Maps etc?

Thanks!

Upvotes: 2

Views: 558

Answers (1)

Paul Jackson
Paul Jackson

Reputation: 2147

In place of select("a") use project("a").by(select("a"))

Upvotes: 4

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