在WHERE子句中使用聚合函数(Neo4j)

Question

如何选择连接到具有多个路径的节点(2)[from]的所有节点？

START from=node(2)
MATCH p=from-->to
where count(p) > 1 
return from,to

致Neo4J团队:是否计划实施Count/Having功能？到目前为止,产品的出色表现!

Answer 1

实际上找到了结合'WITH'关键字的解决方案

START from=node(*) 
MATCH p=from-->to 
WITH from as from , to as to, count(p) as paths 
WHERE paths >1 
RETURN to,paths