如何匹配具有超过 1 个指定关系的节点 [英] How to MATCH nodes with with more than 1 of a specified relationship

问题描述

我试图在 Neo4j 中找到所有 Datum 节点，其中有超过 1 个 GOLDSOURCEFEED 关系指向它.我不在乎什么类型的节点有关系.这是基本查询，您可以查看节点变量等.

I'm trying to find all Datum nodes in Neo4j where there is more than 1 of the GOLDSOURCEFEED relationships pointing to it. And I don't care what type of node has the relationship. Here's the base query just so you can see the node variables etc.

MATCH (p:Datum)<-[:GOLDSOURCEFEED]-()
WHERE exists ( (p:Datum)<-[:GOLDSOURCEFEED]-() )
RETURN p.name

显然，当我放入上面的列表时，我得到了所有具有 GOLDSOURCEFEED 关系的 Datum 节点，我想对其进行细化以仅列出存在多个 GOLDSOURCEFEED 关系的地方.

Obviously when I put in the list above, I get all of the Datum nodes that have a GOLDSOURCEFEED relationship, I want to refine that to only list where there's more than one GOLDSOURCEFEED relationship.

推荐答案

您应该使用有效的度数检查:

You should use an efficient degreeness check:

MATCH (p: Datum)
WHERE SIZE((p)<-[:GOLDSOURCEFEED]-()) > 1 
RETURN p.name;

度数检查只是使用每个节点已经可用的数据，而不需要实际获取路径.当您不关心关系另一端的节点或任何关系属性时，可以执行度数检查.

A degreeness check simply uses data already available for each node, and does not require actually getting the paths. It is possible to perform a degreeness check when you do not care about the node on the opposite end of the relationship or any relationship properties.

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