neo4j - 使用 NEO4j 显示具有多个关系的节点

Question

我使用以下代码创建了一个图表。当每个用户连接到多个视频以及每个视频连接到多个用户时，如何不仅返回节点而且返回关系？

CREATE CONSTRAINT ON (u:User) ASSERT u.user IS UNIQUE;
CREATE CONSTRAINT ON (v:Video) ASSERT v.video IS UNIQUE;

USING PERIODIC COMMIT 100000
LOAD CSV WITH HEADERS FROM 'asdfjkl;' AS line
WITH distinct line.user as user_data
MERGE (:User {user: user_data });

USING PERIODIC COMMIT 100000
LOAD CSV WITH HEADERS FROM 'asdfjkl;' AS line
WITH distinct line.video as video_data
MERGE (:Video {video: video_data });

USING PERIODIC COMMIT 100000
LOAD CSV WITH HEADERS FROM 'asdfjkl;' AS line
MATCH (u:User {user: line.user })
MATCH (v:Video {video: line.video})
MERGE (u)-[:VIEW]->(v);

Answer 1

根据更新的问题进行编辑-获取连接到多个视频的用户的节点和关系:

MATCH (u:User)-[r:VIEW]->(:Video)
WITH u, count(r) AS count
WHERE count > 1
MATCH (u)-[r:VIEW]->(v:Video)
RETURN u,r,v;

获取连接到多个用户的视频的节点和关系:

MATCH (:User)-[r:VIEW]->(v:Video)
WITH v, count(r) AS count
WHERE count > 1
MATCH (u:User)-[r:VIEW]->(v)
RETURN u,r,v;

请注意，在这两种情况下，因为您也需要关系，所以每个 u、r、v 都会得到一行。如果您不想要关系，您可以收集所有视频，例如:

   MATCH (u:User)-[r:VIEW]->(:Video)
   WITH u, count(r) AS count
   WHERE count > 1
   MATCH (u)-[r:VIEW]->(v:Video)
   RETURN u,collect(v);

最后，获取已观看多个视频的用户，其中视频已被多个用户观看:

MATCH (u:User)-[r:VIEW]->(:Video)
WITH u, count(r) AS count
WHERE count > 1
MATCH (u)-[r:VIEW]->(v:Video)<-[r2:VIEW]-(other:User)
RETURN u,r,v,collect(other);