MATCH (c:someNode) WHERE LOWER(c.erpId) contains (LOWER("1"))
OR LOWER(c.constructionYear) contains (LOWER("1"))
OR LOWER(c.label) contains (LOWER("1"))
OR LOWER(c.name) contains (LOWER("1"))
OR LOWER(c.description) contains (LOWER("1"))with collect(distinct c) as rows, count(c) as total
MATCH (c:someNode)-[adtype:OFFICIAL_someNode_ADDRESS]->(ad:anotherObject)
WHERE toString(ad.streetAddress) contains "1"
OR toString(ad.postalCity) contains "1"
with distinct rows+collect( c) as rows, count(c) +total as total
UNWIND rows AS part
RETURN part order by part.name SKIP 20 Limit 20
当我运行以下cypher查询时,它会返回重复的结果。跳过它似乎也不起作用。我在做什么?
答案 0 :(得分:1)
当您使用WITH DISTINCT a, b, c
(或RETURN DISTINCT a, b, c
)时,这意味着您希望每个结果记录({a: ..., b: ..., c: ...}
)都是不同的 - 它不会以任何方式影响内容任何可能属于a
,b
或c
的列表。
以下是可能适合您的简化查询。它根本不使用LOWER()
和TOSTRING()
函数,因为它们似乎是多余的。它也只使用一个MATCH/WHERE
对来查找所有感兴趣的节点。如果WHERE
个节点有true
个{{1}利益。请注意,不需要anotherObject
。
DISTINCT