我目前正在使用以下三个查询从数据库中提取一些信息。
MATCH (u:User { id: "1" }), (x:Label)
WHERE (u)-[:SEES]->(x)
RETURN collect(x) as labels
MATCH (u:User { id: "1" }), (x:Label)
WHERE (u)-[:SEES]->(x)
AND NOT((x)-[:CLASSIFIES]->())
AND NOT(()-[:CLASSIFIES]->(x))
RETURN collect(x) as uniquelabels
MATCH (u:User { id: "1" }), p=(a:Label)-[:CLASSIFIES*]->(b:Label)
WHERE ALL(n IN nodes(path) WHERE ((u)-[:SEES]->(n)))
AND NOT(()-[:CLASSIFIES]->(a))
AND NOT((b)-[:CLASSIFIES]->())
WITH collect(extract(n IN nodes(p) | n.id)) as paths
RETURN distinct paths
将这些查询合并为一个的正确方法是什么?
答案 0 :(得分:1)
首先,我建议不要在MATCH
子句中使用WHERE
语法(ASCII艺术箭头语法),除非您正在执行NOT()
。
我认为你应该可以像这样使用UNION
:
MATCH (u:User { id: "1" })-[:SEES]->(x:Label)
RETURN extract(n IN collect(distinct x) | n.id) as labels
UNION
MATCH (x:Label)
WHERE NOT((x)-[:CLASSIFIES]-())
RETURN extract(n IN collect(distinct x) | n.id) as uniquelabels
UNION
MATCH (u:User { id: "1" }), p=(a:Label)-[:CLASSIFIES*]->(b:Label)
WHERE ALL(n IN nodes(path) WHERE ((u)-[:SEES]->(n)))
AND NOT(()-[:CLASSIFIES]->(a))
AND NOT((b)-[:CLASSIFIES]->())
RETURN DISTINCT collect(extract(n IN nodes(p) | n.id))
可能有一种更优雅的方式,但是,我没有仔细查看查询