Question

我有两个节点，“客户端”和“构建器”。我只想查找客户端只有1个构建器的关系，而该构建器只有一个客户端，所以是1-1关系。到目前为止，我的查询是

MERGE(b:Person{name:csv.name) 
MERGE(c:Person{name:csv.name})
WITH collect(distinct b) as builder, collect(distinct c) as client
UNWIND builder as builders
UNWIND client as clients
WITH builders, clients
WHERE builders = 1 and clients = 1
MATCH (builders:Person)-[bu:builder_for]->(clients:Person)
WITH builders,clients, count(distinct bu) as builds
WHERE builds=1

RETURN distinct builders, clients

这仅返回一对多关系，并且在我的客户列表中仍显示重复项。

The highlighted are the ones I would want to return

更新实施Cybersam非常成功，非常感谢！

Answer 1

假设您要获取所有{em> all builder / client对，它们之间只有一个builder_for关系，此查询将使用{{3 }} COUNT为此：

MATCH (builder:Person)-[rel:builder_for]->(client:Person)
WITH builder, client, COUNT(rel) AS rel_count
WHERE rel_count = 1
RETURN builder, client;

[更新]

如果相反，您想要builder / client对中builder仅具有一个client，反之亦然，则此查询应该起作用：< / p>

MATCH (builder:Person)
WHERE SIZE((builder)-[:builder_for]->()) = 1
MATCH (builder)-[:builder_for]->(client:Person)
WHERE SIZE(()-[:builder_for]->(client)) = 1
RETURN builder, client;

此查询使用有效的关联度检查（在WHERE子句中）以确保builder和client节点分别仅具有单个传出或传入的{{ 1}}关系。

Neo4j一对一关系

1 个答案: