Neo4j：建模多对多关系

Question

我如何为以下数据建模：

“A 人”被“Studio 1”聘请出演“电影 A”
“A 人”被“Studio 1”聘请出演“电影 B”
“A 人”被“Studio 1”聘请出演“电影 C”
“A 人”被“Studio 2”聘请出演“电影 X”
“A 人”被“Studio 2”聘请出演“电影 Y”
“A 人”被“Studio 2”聘请出演“Movie Z”

然后回答问题：“A 人”被“Studio 1”聘请出演了哪些电影？或找到“A 人”出演并被“Studio 1”聘用的电影？

1. CREATE (s1:Studio {name: 'Studio A'}),(s2:Studio {name: 'Studio B'})
2. CREATE (p:Actor {name: 'Person A'})
3. CREATE (m1:Movie {name:'Movie A'}),(m2:Movie {name:'Movie B'}),(m3:Movie {name:'Movie C'})
4. CREATE (m4:Movie {name:'Movie X'}),(m5:Movie {name:'Movie Y'}),(m6:Movie {name:'Movie Z'})

1. MATCH (A:Actor {name: 'Person A'}),(S:Studio {name: 'Studio A'}),(M:Movie {name: 'Movie A'})
 MERGE (S)-[:HIRE]->(A)-[:ACT]->(M)
2. MATCH (A:Actor {name: 'Person A'}),(S:Studio {name: 'Studio A'}),(M:Movie {name: 'Movie B'}) 
MERGE (S)-[:HIRE]->(A)-[:ACT]->(M)
3. MATCH (A:Actor {name: 'Person A'}),(S:Studio {name: 'Studio A'}),(M:Movie {name: 'Movie C'}) 
MERGE (S)-[:HIRE]->(A)-[:ACT]->(M)

4. MATCH (A:Actor {name: 'Person A'}),(S:Studio {name: 'Studio B'}),(M:Movie {name: 'Movie X'})
 MERGE (S)-[:HIRE]->(A)-[:ACT]->(M)
5. MATCH (A:Actor {name: 'Person A'}),(S:Studio {name: 'Studio B'}),(M:Movie {name: 'Movie Y'}) 
MERGE (S)-[:HIRE]->(A)-[:ACT]->(M)
6. MATCH (A:Actor {name: 'Person A'}),(S:Studio {name: 'Studio B'}),(M:Movie {name: 'Movie Z'}) 
MERGE (S)-[:HIRE]->(A)-[:ACT]->(M)

Answer 1

问题是，使用当前模型，您将失去关于工作室聘请演员在哪部电影中表演的上下文。想象一下，如果两个工作室聘请一个人在两部电影中表演。你最终会得到的是：

为了不丢失工作室雇用演员的电影的上下文，您可以引入一个中间节点作业，例如：

有了这个图形模型，您就不会忘记工作室雇用此人出演的电影的上下文。如果您需要一些有关基本 Cypher 查询的帮助，请查看 Cypher 培训课程：https://neo4j.com/graphacademy/training-querying-40/enrollment/

Neo4j：建模多对多关系

Neo4j : Modeling Many to Many Relation

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