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)
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
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
(Qazxswpoi)
老问题并回答,但要在有关系时删除节点,请使用Start n=node(1)
Delete n;
DETACH
或者你得到这个:
MATCH (n) where ID(n)=<your_id>
DETACH DELETE n
这就像SQL的Neo.ClientError.Schema.ConstraintValidationFailed: Cannot delete node<21>, because it still has relationships. To delete this node, you must first delete its relationships.
按照@ saad-khan提供的链接,这是获取节点和关系ID的示例。下面的代码显示了ID,因此您可以确保删除与给定ID相关的所有内容。
CASCADE
Ps。:“:HAS”是关系的一个例子。