Gremlin.Net一次遍历即可获得边线Id,InV,OutV和属性

时间:2018-08-16 16:15:18

标签: c# .net gremlin tinkerpop

当前,只需一次遍历,我就可以做到:

Edge edge = g.E().Next();
var inv = edge.InV;
var outv = edge.OutV;
var id = edge.Id;

这使我可以获得边缘的ID,以及边缘之间的顶点的ID。或者,我可以这样做:

IDictionary<object, object> dict = g.E().ValueMap<object, object>(true).Next();
var id = dict[T.id]
var edgeProp = dict["$edgePropertyName"];

这允许我获取属性和ID,但不能获取边缘的ID。有没有办法在单个遍历中同时获得顶点和属性?

2 个答案:

答案 0 :(得分:3)

好的,只需使用project()

gremlin> g.E().
......1>   project('id','properties','out','in').
......2>     by(id).
......3>     by(valueMap()).
......4>     by(outV().id()).
......5>     by(inV().id())
==>[id:7,properties:[weight:0.5],out:1,in:2]
==>[id:8,properties:[weight:1.0],out:1,in:4]
==>[id:9,properties:[weight:0.4],out:1,in:3]
==>[id:10,properties:[weight:1.0],out:4,in:5]
==>[id:11,properties:[weight:0.4],out:4,in:3]
==>[id:12,properties:[weight:0.2],out:6,in:3]

答案 1 :(得分:0)

将接受的答案翻译成C#:

var result = g.E()
    .Project<object>("id", "properties", "out", "in")
    .By(__.Id())
    .By(__.ValueMap<string, object>())
    .By(__.OutV().Id())
    .By(__.InV().Id())
    .Next();