在SQL中:
Delete From Person Where ID = 1;
在Cypher中,通过ID删除节点的脚本是什么?
(编辑: ID =Neo4j的内部节点ID)
发布于 2015-01-26 12:40:29
假设您引用的是Neo4j的内部节点id:
MATCH (p:Person) where ID(p)=1
OPTIONAL MATCH (p)-[r]-() //drops p's relations
DELETE r,p
如果你在节点上引用你自己的属性'id‘:
MATCH (p:Person {id:1})
OPTIONAL MATCH (p)-[r]-() //drops p's relations
DELETE r,p
发布于 2016-09-21 04:11:56
发布于 2019-04-03 00:41:58
老问题和答案,但是要在节点有关系时将其删除,请使用DETACH
MATCH (n) where ID(n)=<your_id>
DETACH DELETE n
否则你会得到这样的结果:
Neo.ClientError.Schema.ConstraintValidationFailed: Cannot delete node<21>, because it still has relationships. To delete this node, you must first delete its relationships.
它就像SQL的CASCADE
。
https://stackoverflow.com/questions/28144751
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