[英]Finding nodes that do not have specific relationship (Cypher/neo4j)
I have a neo4j db with the following: 我有一个neo4j数据库,具有以下内容:
a:Foo
b:Bar
about 10% of db have (a)-[:has]->(b)
大约10%的db有
(a)-[:has]->(b)
I need to get only the nodes that do NOT have that relationship! 我只需要获得那些没有这种关系的节点!
previously doing ()-[r?]-()
would've been perfect! 以前做
()-[r?]-()
会很完美! However it is no longer supported :( instead, doing as they suggest a 然而,它不再受支持:(而是,正如他们建议的那样
OPTIONAL MATCH (a:Foo)-[r:has]->(b:Bar) WHERE b is NULL RETURN a
gives me a null result since optional match needs BOTH nodes to either be there or BOTH nodes not to be there... 给我一个null结果,因为可选匹配需要BOTH节点在那里或BOTH节点不在那里......
So how do i get all the a:Foo
nodes that are NOT attached to b:Bar
? 那么我怎么得到所有的
a:Foo
没有附加到b:Bar
a:Foo
节点?
Note: dataset is millions of nodes so the query needs to be efficient or otherwise it times out. 注意:数据集是数百万个节点,因此查询需要高效或以其他方式超时。
那就是
MATCH (a:Foo) WHERE not ((a)-[:has]->(:Bar)) RETURN a;
如果您正在寻找所有单身人士/孤儿,这也有效:
MATCH (a:Foo) WHERE not ((a)--()) RETURN a;
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