基于neo4j中每个节点的祖先属性按关系匹配节点

在对象部件具有对象部件的部件的场景中：

组件1

A:Object{name:ABC} is PART_OF B:Object{name:DEF} is PART_OF C:Object{name:GHI}

组件2

D:Object{name:JKL} is PART_OF E:Object{name:MNO} is PART_OF F:Object{name:PQR} is PART_OF G:Object{name:STU}

组件1的某个组件对象连接到组件2

C is JOINED_TO G

如何

匹配

部件1中部件链的任何部分具有给定名称特性的所有对象，这些对象与任何部分具有给定名称特性的对象相连接

我的猜测是：

MATCH ob1:Object{name:DEF} - [:PART_OF*] -> ob2:Object - [:JOINED_TO] - ob3:Object <- [:PART_OF*] - ob4:Object{name:PQR}
RETURN ob2, ob3

---

我用它来创建示例图：

CREATE (A:Object{name:'ABC'})-[:PART_OF]->(B:Object{name:'DEF'})-[:PART_OF]->(C:Object{name:'GHI'}),
(D:Object{name:'JKL'})-[:PART_OF]->(E:Object{name:'MNO'})-[:PART_OF]->(F:Object{name:'PQR'})-[:PART_OF]->(G:Object{name:'STU'})
CREATE (C)-[:JOINED_TO]->(G)

我用这个来查询图表：

MATCH (ob1:Object{name:'DEF'}) - [:PART_OF*] -> (ob2:Object) - [:JOINED_TO] - (ob3:Object) <- [:PART_OF*] - (ob4:Object{name:'PQR'})
RETURN ob2, ob3

---

MATCH（ob1:Object{name:'DEF}）-[：PART_OF*]->（ob2:Object）-[：JOINED_TO]-（ob3:Object）显示的内容应该会得到（许多，可能重复的）结果行。（但是，您的查询没有指定任何名称属性）。您能提供一些密码来创建显示您的问题的示例数据吗？我将检查并接受。结果中的C&G是完整对象的简写。令我欣慰的是，你真正的密码和我粗糙的伪密码是多么的相似。
MATCH (ob1:Object{name:'DEF'}) - [:PART_OF*] -> (ob2:Object) - [:JOINED_TO] - (ob3:Object) <- [:PART_OF*] - (ob4:Object{name:'PQR'})
RETURN ob2, ob3