[英]Neo4j query to ignore parent nodes which doesn't satisfy a condition but keep the same structure
[英]Neo4j Query - Find all nodes which satisfy a property condition and have relationship
我有一个成功运行的查询:
match (n:A {tag_no:"N2203"})<-[:rel_a]-(v:B)-[:rel_b]->(r)<-[:rel_c]-(n) WHERE (v.invoice_date>="2012-08-01" AND v.invoice_date<"2016-02-01") with v, collect(r) as rs where all (x in rs where x.date<"2016-08-01") return count(v) as count;
我需要进一步过滤此查询。
我需要r nodes with max(r.date) for each v
关联r nodes with max(r.date) for each v
,并找出其中有多少个节点与另一个D
类型的节点有关系
我正在尝试此查询,它引发语法错误
match (n:A {tag_no:"N2203"})<-[:rel_a]-(v:B)-[:rel_b]->(r)<-[:rel_c]-(n) WHERE (v.invoice_date>="2012-08-01" AND v.invoice_date<"2016-02-01") with v, max(r.date) as date collect(r) as rs where (all (x in rs where x.date<"2016-08-01") AND filter(x in rs where x.date=date)[0]<-[:rel_c]-(d:D)) return count(v) as count;
我还尝试了许多其他组合,但是都抛出了一些语法错误。 感谢所有帮助。
主要问题是查询解析器现在无法确定filter(x in rs where x.date=date)[0]
确实是一个节点,因此语法中不允许这样做。 幸运的是,有可能以一些冗长的代价解决此问题:
WITH
子句为节点变量引入别名( n
)。 这将允许您使用WHERE <pattern>
语法。 WHERE
子句中引入新变量,因此只需使用(:D)
而不是(d:D)
。 因此查询将如下所示(显然,我尚未对其进行测试):
MATCH (n:A {tag_no:"N2203"})<-[:rel_a]-(v:B)-[:rel_b]->(r)<-[:rel_c]-(n)
WHERE v.invoice_date>="2012-08-01"
AND v.invoice_date<"2016-02-01"
WITH
v,
max(r.date) AS date,
collect(r) AS rs
WHERE all(x IN rs WHERE x.date<"2016-08-01")
WITH filter(x in rs where x.date=date)[0] AS n, v
WHERE (n)<-[:rel_c]-(:D)
RETURN count(v) AS count;
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