用Gremlin递归查询更简单的树状结构 [英] Recursively query simpler tree-like structures with Gremlin

问题描述

请考虑以下数据:

g.addV('RootTopic').property('name', 'A').as('A')
.addV('RootTopic').property('name', 'M').as('M')
.addV('Topic').property('name', 'A1').as('A1')
.addV('Topic').property('name', 'A2').as('A2')
.addV('Topic').property('name', 'B1').as('B1')
.addV('Topic').property('name', 'B2').as('B2')
.addV('Topic').property('name', 'N1').as('N1')
.addV('Topic').property('name', 'N2').as('N2')
.addV('Topic').property('name', 'O1').as('O1')
.addE('refines').from('A').to('A1')
.addE('refines').from('A').to('A2')
.addE('refines').from('A1').to('B1')
.addE('refines').from('A1').to('B2')
.addE('refines').from('M').to('N1')
.addE('refines').from('M').to('N2')
.addE('refines').from('N2').to('O1')
.addE('refines').from('N2').to('O2')

我想要的是使用tree()步骤得到的东西:

What I would like is something that one gets by using the tree()-step:

g.V().hasLabel('RootTopic').repeat(out()).times(2).emit().tree()

但是，这会拉出整个顶点.在这种情况下，我真正需要的是顶点的属性，例如名称，这样我们就可以得到一棵包含例如只是顶点的名称属性.

However, this pulls out the full vertex. What I really just need in this situation are properties of the vertex, e.g. the name, such that we get a tree that contains e.g. just the name-property of the Vertex.

我知道，如果我写.tree().by('name')，我似乎得到了一个以名称为键的树，但是我试图找到一种方法，使我可以选择例如顶点的多种属性，例如只是具有某些特定元属性的某种属性.

I know that if I write .tree().by('name') I seem to get a tree with the name as key, but I am trying to find a way which would allow me to select e.g. multiple properties of a Vertex, or e.g. just a certain property having some specific meta-property.

这可能吗?

推荐答案

by()调制器可以将多个属性键值用作参数.您还可以传入匿名遍历，从而允许:

The by() modulator can take more than just a property key value as an argument. You can also pass in an anonymous traversal which would thus allow:

g.V().hasLabel('RootTopic').
  repeat(out()).times(2).
    emit().
  tree()
    by(values('name','k1','k2').fold())

，否则，如果输出更为复杂，则可以使用project():

or you might use project() if you had more complex output:

g.V().hasLabel('RootTopic').
  repeat(out()).times(2).
    emit().
  tree()
    by(project('name','k1','degree').
         by('name').
         by('k1').
         by(both().count())

这里要讲的要点是，通过匿名遍历，您可以开发几乎任何想要的输出.

The main point to take away here is that with an anonymous traversal you can develop just about any output you would like.

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