什么是通过ID删除节点的Cypher脚本？

Question

在SQL中:

Delete From Person Where ID = 1;

在Cypher中,按ID删除节点的脚本是什么？

(已编辑:ID = Neo4j的内部节点ID)

Answer 1

假设你指的是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

如果您在节点上引用自己的属性"id":

 MATCH (p:Person {id:1})
 OPTIONAL MATCH (p)-[r]-() //drops p's relations
 DELETE r,p

Answer 2

对于id为"xx"的节点,最干净的扫描是

MATCH(n)其中id(n)= xx DETACH DELETE n

(https://neo4j.com/docs/developer-manual/current/cypher/#delete-delete-all-nodes-and-relationships)

Answer 3

老问题和答案，但是要在有关系时删除节点，请使用DETACH

MATCH (n) where ID(n)=<your_id> 
DETACH DELETE n

否则你会得到这个：

Neo.ClientError.Schema.ConstraintValidationFailed: Cannot delete node<21>, because it still has relationships. To delete this node, you must first delete its relationships.

这就像SQL的CASCADE