Question

我有以下Neo4j场景（我正在使用Neo4j 2.0.1）：

节点：

User
Token

关系：

friends_with -> bidirectional relationship between Users
is_authorized_by -> directed relationship between a User and a Token

对于每个用户，我需要获得其令牌（如果它具有有效的令牌）或有效的朋友的令牌。

我写了以下Cypher查询：

MATCH (user:User) WHERE user.id IN ['123', '456'] 
OPTIONAL MATCH (user)-[:is_authorized_by]->(user_token:Token) 
    WHERE HAS (user_token.access_token) AND timestamp()/1000 < user_token.expiration_timestamp 
OPTIONAL MATCH (user)-[:friends_with*1..3]-(friend:User)-[:is_authorized_by]->(friend_token:Token) 
    WHERE HAS (friend_token.access_token) AND timestamp()/1000 < friend_token.expiration_timestamp  
RETURN user.id as userId, coalesce(user_token.access_token, friend_token.access_token) AS token

当然，这不起作用，因为第二个OPTIONAL MATCH不仅匹配第一个朋友的令牌。

我想做的是：

MATCH (user:User) WHERE user.id IN ['123', '456'] 
OPTIONAL MATCH (user)-[:is_authorized_by]->(user_token:Token) 
    WHERE HAS (user_token.access_token) AND timestamp()/1000 < user_token.expiration_timestamp 
OPTIONAL MATCH (user)-[:friends_with*1..3]-(friend:User)-[:is_authorized_by]->(friend_token:Token) 
    WHERE HAS (friend_token.access_token) AND timestamp()/1000 < friend_token.expiration_timestamp  <--- limit this to 1
RETURN user.id as userId, coalesce(user_token.access_token, friend_token.access_token) AS token

如何实现这一目标？

修改1：

我只对直接朋友的代币感兴趣：

MATCH (user:User) WHERE user.id IN ['123', '456'] 
OPTIONAL MATCH (user)-[:is_authorized_by]->(user_token:Token) 
    WHERE HAS (user_token.access_token) AND timestamp()/1000 < user_token.expiration_timestamp 
OPTIONAL MATCH (user)-[:friends_with]-(friend:User)-[:is_authorized_by]->(friend_token:Token) 
    WHERE HAS (friend_token.access_token) AND timestamp()/1000 < friend_token.expiration_timestamp  
RETURN user.id as userId, coalesce(user_token.access_token, friend_token.access_token) AS token

Answer 1

要从返回的列表中返回任何一个，您可以应用MAX等聚合函数，即

将user.id作为userId，MAX返回（合并（user_token.access_token， friend_token.access_token））AS令牌

Answer 2

由于coalesce()函数最多只返回一个access_token，因此您只想每userId返回一行：

RETURN DISTINCT user.id as userId, coalesce(user_token.access_token, friend_token.access_token) AS token

Answer 3

这会返回多少行？你想把它限制在一个吗？

OPTIONAL MATCH (user)-[:is_authorized_by]->(user_token:Token)

你想要每个朋友一个令牌还是只需要一个朋友令牌？

您的令牌似乎也没有连接？所以你可能也想要做这样的事情？只是猜测，不确定这是否会像你一样表现。

MATCH (user:User {id:'123'})
WITH user, head(filter(token in 
          extract(p in (user)-[:is_authorized_by]->(:Token) | last(nodes(p))) 
          WHERE token.access_token AND timestamp()/1000 < token.expiration_timestamp)) as user_token
OPTIONAL MATCH (user)-[:friends_with*1..3]-(friend:User)
WITH user, user_token, 
     head(filter(token in 
          extract(p in (friend)-[:is_authorized_by]->(:Token) | last(nodes(p))) 
          WHERE token.access_token AND timestamp()/1000 < token.expiration_timestamp)) as friend_token
RETURN user.id as userId, coalesce(user_token.access_token, friend_token.access_token) AS token

限制Neo4j Cypher中的匹配数量

3 个答案: