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!
我只需要获得没有这种关系的节点！
previously doing ()-[r?]-() would've been perfect! However it is no longer supported :( instead, doing as they suggest a
以前做（） - [r？] - （）会很完美！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：没有附加到b：Bar的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。
2 个解决方案
#1
24
That would be
那就是
MATCH (a:Foo) WHERE not ((a)-[:has]->(:Bar)) RETURN a;
#2
3
This also works if you're looking for all singletons/orphans:
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;
#1
24
That would be
那就是
MATCH (a:Foo) WHERE not ((a)-[:has]->(:Bar)) RETURN a;
#2
3
This also works if you're looking for all singletons/orphans:
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;