Cypher:如果关系不存在，如何返回节点? [英] Cypher: How to return a node if the relationship doesn't exist?

问题描述

我有节点的用户"和建筑物"标签，关系为User-[:HAS_PERMISSION]->Building

I have User and Building labels for nodes, the relationships are User-[:HAS_PERMISSION]->Building

要获取用户有权访问的所有建筑物，请使用MATCH (u:User {id: {id}}),(u)-[r:HAS_PERMISSION]->(b:Building) return b

To get all buildings that a user is having permissions to so I use MATCH (u:User {id: {id}}),(u)-[r:HAS_PERMISSION]->(b:Building) return b

如果用户存在并且没有关系，我该如何返回用户? (基本上，我也想知道用户是否存在.)

How can I just return the user in case he exists and there are no relationships? (basically I also want to know if the user exists at all..)

推荐答案

不确定我是否正确理解了您的问题，但我认为您需要OPTIONAL MATCH.该查询显示所有有权访问或不访问建筑物的用户.

Not sure if I understand your question correctly, but I think you'll need an OPTIONAL MATCH. This query shows all users with or without access to a building.

MATCH (u:User)
OPTIONAL MATCH (u)-[r:HAS_PERMISSION]->(b:Building) 
RETURN u, b

或者您的要求可能更简单.如果只需要所有用户，则无需使用模式匹配:

Or maybe your requirement is simpler. If you just want all users, there's no need to use pattern matching:

MATCH (u:User)
RETURN u

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