问题描述
我有一个neo4j数据库与以下：
I have a neo4j db with the following:
a:Foo
b:Bar
约有10％的db has（a） - [：has] - >（b）
about 10% of db have (a)-[:has]->(b)
我需要只获得没有这种关系的节点！
I need to get only the nodes that do NOT have that relationship!
以前做的（） - [r？] - 已经完美！但是它不再支持:(相反，做他们建议一个可选的匹配（a：Foo） - [r：has] - >（b：Bar）WHERE b是NULL返回a匹配需要BOTH节点要么在那里，要么BOTH节点不在那里...
previously doing ()-[r?]-() would've been perfect! however it is no longer supported :( instead, doing as they suggest a "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...
那么如何获得所有的a：Foo到b：Bar？
So how do i get all the "a:Foo" nodes that are NOT attached to "b:Bar"?
注意：数据集是数百万个节点，因此查询需要有效率，否则超时。
Note: dataset is millions of nodes so the query needs to be efficient or otherwise it times out.
谢谢！
推荐答案
这将是
MATCH (a:Foo) WHERE not ((a)-[:has]->(:Bar)) RETURN a;
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