neo4j：如何 return 带有节点和关系标签的路径？

Question

我的查询 returns 路径：

match path = ...
return path

但缺少节点和关系标签。我怎样才能添加它们？

return path, labels(path)

报错

Answer 1

您可以在整个路径中使用命名节点。然后对每个命名节点 (labels(a) + labels(b)) 的标签求和。在它之后，您可以使用 UNWIND 和 collect distinct 删除重复项。此外，您可以执行相同的操作来获取关系类型，但使用 relationships() 函数。

match path = (a)-->(b)
with path, labels(a) + labels(b) as labels, relationships(path) as relationships
unwind(labels) as label
unwind(relationships) as relationship
return path, collect(distinct label) as labels, collect(distinct type(relationship)) as relationships

Answer 2

我取得的最好成绩是

match path = ...
return path, extract (x in rels(path) | type(x)) as types, extract (n in nodes(path) | labels(n)) as labels

它允许在后端重建整个路径结构。虽然仍然包含很多重复

neo4j：如何 return 带有节点和关系标签的路径？

neo4j: how to return paths with labels of nodes and relations?

neo4j

graph-databases

cypher