Question

(client)-[:listens]->(album)-[:contains]->(song)是我的关系。我需要找到有多少不同的歌曲和客户听过哪首歌已经听过ID为4的歌曲（f.e。）。

在SQL中，我可以通过双嵌套JOIN和SELECT轻松实现这一点，但由于我是Cypher的新手，我有点迷茫。方案如下：

查找哪些客户（他们的ID）已收听ID为4的歌曲
从上一步中找到的客户获取所有已收听的歌曲，并将其返回（以及他们的计数）。

我试着像smth一样开始：

MATCH (c:Client)--(a:Album)--(s:Song)
WITH a, s, {cIDs: c.clientID} AS clientsIDs
// MATCH that gets all songs in albums listened by clients WHERE c.clientID IN clientsIDs -- how to make that?
WITH COUNT(s), COLLECT(s) as songs
WHERE s.songID = "4"
RETURN songs

这是另一次尝试：

MATCH (c:Client)--(a:Album)--(s:Song)
WHERE s.songID = "4"
WITH a, s, {cID: c.clientID} AS clientsIDs
MATCH (c {clientID: clientsIDs})-[:LISTENS]->(a)-[:CONTAINS]->(s)
WITH COUNT(s) AS songsCount, COLLECT(s) as songs
RETURN songs, songsCount

显然不起作用。

我无法绕过将clientsIDs传递给另一个MATCH声明的部分......

Answer 1

我认为你可能会过度复杂化。这是我的努力......

// find all of the clients that have listed to the song with id =4
MATCH (c:Client)--(a:Album)--(s:Song)
WHERE s.songID = "4"

// then with those clients find all of the songs they have listened to
WITH c 
MATCH (c)-[:LISTENS]->(a)-[:CONTAINS]->(s)
WITH COLLECT(s) as songs
RETURN songs, size(songs) as songsCount

cypher中的嵌套MATCH语句

1 个答案: