如何枚举通过Cypher返回的路径上的节点和关系 [英] How to enumerate nodes and relationships along path returned via Cypher

问题描述

我在这里打开了一个问题:如何查找Neo4j中的特定子图，使用where子句查找特定条件的路径.然而，当我尝试做类似获取关系类型的事情时，我却无法做到.

I opened this question here: How to find specific subgraph in Neo4j using where clause to find a path of a certain criteria. Yet when I try to do things like get the relationship type I cannot.

例如，我尝试了MATCH p = (n:Root)-[rs1*]->() WHERE ALL(rel in rs1 WHERE rel.relevance is null) RETURN nodes(p), TYPE(relationships(p))

For example I tried MATCH p = (n:Root)-[rs1*]->() WHERE ALL(rel in rs1 WHERE rel.relevance is null) RETURN nodes(p), TYPE(relationships(p))

但是我得到了错误:

Type mismatch: expected Relationship but was Collection<Relationship>

我认为我需要使用WITH子句，但不确定.

I think I need to use a WITH clause but not sure.

类似地，我想要一个节点的ID，但这也失败了.

Similarly I wanted the ID of a node but that also failed.

推荐答案

问题是relationships返回一个集合，而type函数仅适用于单个关系.解决此问题的方法主要有两种.

The problem is that relationships returns a collection and the type function only works on a single relationship. There are two main approaches to solve this.

使用 UNWIND 可以单独获取每个关系的行:

Use UNWIND to get a separate row for each relationship:

MATCH p = (n:Root)-[rs1*]->()
WHERE ALL(rel in rs1 WHERE rel.relevance is null)
WITH relationships(p) AS rs
UNWIND n, rs AS r
RETURN n, type(r)

使用 extract 获得结果列表中(每个根节点在单行中):

Use extract to get the results in a list (in a single row per root node):

MATCH p = (n:Root)-[rs1*]->()
WHERE ALL(rel in rs1 WHERE rel.relevance is null)
WITH n, relationships(p) AS rs
RETURN n, extract(r IN rs | type(r))

或更短:

MATCH p = (n:Root)-[rs1*]->()
WHERE ALL(rel in rs1 WHERE rel.relevance is null)
RETURN n, extract(r IN relationships(p) | type(r))

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