sorting - Neo4j 密码复杂查询排序，计数，收集前求和

Question

我是 neo4j db 的新手，刚开始学习它，正在寻求帮助，因为我被卡住了。是否有可能在一个密码查询中得到它？怎么办？

我的图结构是这样的:

(s:Store)-[r:RELEASED]->(m:Movie)<-[r1:ASSIGNED]-(cat:MovieCategorie)

我如何获得这些数据？

• 电影店(得到)
• 电影(得到)
• 该商店中最常见的 5 类电影(在使用 collect(cat.name)[0..5] 之前我不知道如何对它们进行排序)

任何人都可以建议如何获取这些数据？我尝试了很多次但都失败了，这就是我得到的但它不起作用。

match (s:Store) 
with s
match (s)-[r:RELEASED]->(m:Movie)
with s,m 
match (m)<-[r1:ASSIGNED]-(cat:MovieCategorie)
with s, m, count(r1) as stylesCount, cat
order by stylesCount
return distinct s as store, collect(cat.name)[0..5] as topCategories
order by store.name

谢谢!

---

好的，当我正确查询并进一步开发此查询时，通过组合多个聚合函数 COUNT 和 SUM 遇到了一些问题。

我的查询可以很好地找到每个商店的前 5 个类别:

MATCH (s:Store)
OPTIONAL MATCH (s)-[:RELEASED]->(m:Movie)<-[r:ASSIGNED]-(cat:MovieCategorie)
WITH s, COUNT(r) AS count, cat
ORDER BY  count DESC
RETURN c AS Store, COLLECT(distinct cat.name) AS `Top Categories`
ORDER BY Store.name

在此查询之上，我需要计算这家商店有多少浏览量 sum(m.viewsCount) 作为总商店浏览量。我尝试添加到与 COUNT 相同的 WITH 语句中，并尝试将其作为返回，在这两种情况下，它都无法正常工作。有什么建议，例子吗？我仍然对 WITH 和聚合函数的工作原理感到困惑......:(

创建示例数据库

CREATE (s1:Store) SET s1.name = 'Store 1'
CREATE (s2:Store) SET s2.name = 'Store 2'
CREATE (s3:Store) SET s3.name = 'Store 3'

CREATE (m1:Movie) SET m1.title = 'Movie 1', m1.viewsCount = 50
CREATE (m2:Movie) SET m2.title = 'Movie 2', m2.viewsCount = 50
CREATE (m3:Movie) SET m3.title = 'Movie 3', m3.viewsCount = 50
CREATE (m4:Movie) SET m4.title = 'Movie 4', m4.viewsCount = 50
CREATE (m5:Movie) SET m5.title = 'Movie 5', m5.viewsCount = 50

CREATE (c1:MovieCategorie) SET c1.name = 'Cat 1'
CREATE (c2:MovieCategorie) SET c2.name = 'Cat 2'
CREATE (c3:MovieCategorie) SET c3.name = 'Cat 3'

CREATE (m1)<-[:ASSIGNED]-(c1)
CREATE (m1)<-[:ASSIGNED]-(c3)
CREATE (m2)<-[:ASSIGNED]-(c2)
CREATE (m3)<-[:ASSIGNED]-(c1)
CREATE (m3)<-[:ASSIGNED]-(c2)
CREATE (m3)<-[:ASSIGNED]-(c3)
CREATE (m4)<-[:ASSIGNED]-(c1)
CREATE (m4)<-[:ASSIGNED]-(c3)
CREATE (m5)<-[:ASSIGNED]-(c3)

CREATE (s1)-[:RELEASED]->(m1)
CREATE (s1)-[:RELEASED]->(m3)
CREATE (s1)-[:RELEASED]->(m4)
CREATE (s1)-[:RELEASED]->(m5)

CREATE (s2)-[:RELEASED]->(m1)
CREATE (s2)-[:RELEASED]->(m2)
CREATE (s2)-[:RELEASED]->(m3)
CREATE (s2)-[:RELEASED]->(m4)
CREATE (s2)-[:RELEASED]->(m5)

CREATE (s3)-[:RELEASED]->(m1)

---

解决了!!最后我做到了!技巧是在一切之后再使用一场比赛，太棒了——现在我可以安心 sleep 了。谢谢。

MATCH (s:Store)-[:RELEASED]->(m:Movie)<-[r:ASSIGNED]-(cat:MovieCategorie)
with  s,count(r) as catCount,  cat
order by catCount desc
with s, collect( distinct cat.name)[0..5] as TopCategories
match (s)-[:RELEASED]->(m:Movie)
return s as Store, TopCategories, sum(m.viewsCount) as TotalViews

Answer 1

好的，太快了 :D 我终于明白了!

match (s:Store) 
with s
match (s)-[r:PUBLISHED]->(m:Movie)
with s
match (s)<-[r2:ASSIGNED]-(cat:MovieCategorie)
with s, count(r2) as stylesCount, cat
order by stylesCount desc
return distinct s,  collect(distinct cat.name)[0..5] as topCategories
order by s.name

所以技巧是首先在 with 中使用 count()，然后按 with 排序，并收集 DISTINCT 作为返回。我不太确定这些 multiple with 语句，会尝试清理它。 ;)