Graphdb/Neo4j与另一个关系的关系，或与3个节点的关系 [英] Graphdb/Neo4j relationship to another relationship, or relationship with 3 nodes

问题描述

我想在Neo4j中制作IMDB的角色/角色结构.我需要标签Person，Movie和Character. Character，因为一个角色可以在多部电影中由不同的人扮演.

I want to make IMDB's characters/roles structure in Neo4j. I'll need labels Person, Movie and Character. Character, because a character can be in multiple movies, played by different people.

没有Character，这很容易:

(Person)-[:PLAYS_IN]->(Movie)

但是PLAYS_IN是Character，因此它类似于:

But PLAYS_IN is a Character, so it would be something like:

(Person)-[:PLAYS_AS]->(Character)-[:PLAYS_IN]->(Movie)

，但这不起作用，因为它没有直接的Person-Movie关系.没有这个，所有曾经扮演彼得·帕克的人都会出现在每部拥有彼得·帕克的电影中.

but that doesn't work, because it doesn't have a direct Person-Movie relationship. Without that, everyone who ever played Peter Parker, is in every movie that has a Peter Parker.

必须存在一个人与电影的关系，但也必须有一个人与电影-字符的关系.如何?这可能有效，但这很讨厌:

There must be a Person-Movie relationship, but also a Person-Movie-Character relationship. How? This could work, but that's just nasty:

(Person)-[:PLAYS_IN {uuid}]->(Movie), (Character {uuid})

因为现在我正在创建自己的外键类型的关系.那是非常ungraphdb的.但这有效:

because now I'm creating my own foreign key kind of relationship. That's very ungraphdb. But it works:

MATCH (p:Person)-[r:PLAYS_IN]->(m:Movie), (c:Character)
WHERE c.uuid = r.uuid
RETURN p, c, m

通过构建笛卡尔乘积=(这是非常RDBMS，但不是非常的graphdb.而且我无法查询Character-Movie或Character-Person，因为这不是真正的关系.

by building a cartesian product =( which is very RDBMS, but not very graphdb. And I can't query Character-Movie or Character-Person, because that's not a real relationship.

如何在Neo4j中使用3个外键(movie_id, character_id, person_id)创建RDBMS链接表?

How do I make a RDBMS link table with 3 foreign keys (movie_id, character_id, person_id) in Neo4j??

修改1
RDBMS等效项:

edit 1
The RDBMS equivalent:

movies (id, title)                         # e.g. Dragon Tattoo, or Spider's Web
people (id, name)                          # e.g. Rooney Mara, or Claire Foy
characters (id, name)                      # e.g. Lisbeth Salander
roles (movie_id, person_id, character_id)  # 2 rows with 1 distinct character_id

推荐答案

超图.但是由于neo4j不支持超图，因此您需要对其建模.例如:

Your problem is solved by hypergraphs. But since neo4j does not support hypergraphs, you need to model them. For example:

CREATE (P1:Person {name: 'Tobey Maguire'})
CREATE (P2:Person {name: 'Nicholas Hammond'})
CREATE (CW1:CreativeWork {name: 'Spider-Man'})
CREATE (CW2:CreativeWork {name: 'The Amazing Spider-Man (TV Series)'})
CREATE (CH:Character {name: 'Spider-Man'})

CREATE (A1:Role)
CREATE (A2:Role)

CREATE (P1)-[:PLAYS_AS]->(A1)
CREATE (A1)-[:HAS_CHARACTER]->(CH)
CREATE (A1)-[:PLAYS_IN]->(CW1)

CREATE (P2)-[:PLAYS_AS]->(A2)
CREATE (A2)-[:HAS_CHARACTER]->(CH)
CREATE (A2)-[:PLAYS_IN]->(CW2)

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