Question

我想要那些名称不在a_row集合中的节点。例如：b,c拥有b_row，将a_row与b_row b,c,e合并后，entity.name正在a,b,c,d，a,d正在match (a)-[r:relations]->(b) with collect(distinct a.FirstName) as a_row match (a)-[r:relations]->(b) with a_row + collect(distinct b.LastName) as b_row match (n:entity) with b_row where n.name <> b_row return n。所以我希望节点形式的输出0xF不在列表表单中。

0b1111

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Answer 1

尝试按IN operator更改<>。这样：

match (a)-[r:relations]->(b) with collect(distinct a.FirstName) as a_row
match (a)-[r:relations]->(b) with a_row + collect(distinct b.LastName) as b_row
match (n:entity) where n.name NOT IN b_row return n

Answer 2

第一

match (a)-[r:relations]->(b) with collect(distinct a.FirstName) as a_row
match (a)-[r:relations]->(b) with a_row + collect(distinct b.LastName) as b_row
match (n:entity) with b_row where n.name <> b_row return n

无效，因为最后with会丢弃n，并且可以简化为

MATCH (a)-[r:relations]->(b) 
WITH collect(distinct a.FirstName)+collect(distinct b.LastName) as b_row
MATCH (n:entity)
// b_row in now a list, so if n.name is a string, <> is always false
WHERE n.name NOT IN b_row 
RETURN n

我将简化为

MATCH (a)-[r:relations]->(b) 
WITH collect(distinct a.FirstName)+collect(distinct b.LastName) as b_row
MATCH (n:entity)
RETURN n, b_row

首先验证您的查询是否在过滤之前提升了您的想法。

获取未收集的节点

2 个答案: