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我是 neo4j db 的新手，刚开始学习它，寻求帮助，因为我被卡住了。是否可以在一个密码查询中获得它？如何？

我的图形结构如下所示：

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

我怎么能得到这些数据？

  • 电影商店（有）
  • 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，所以当我的查询正确并且我正在进一步开发这个查询时，通过组合多个聚合函数 COUNT 和 SUM 遇到了一些问题。

我的查询女巫非常适合查找每个商店的前 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

在这个查询之上，我需要计算这家商店有多少浏览量 sum(m.viewsCount) 作为商店浏览总数。我尝试添加与 COUNT 相同的 WITH 语句，并尝试将其作为回报，在这两种情况下，它都无法按我的意愿工作。有什么建议、例子吗？我仍然对 WITH 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！！终于我做到了！诀窍是在所有事情之后再使用一场比赛，太好了 - 现在我可以安然入睡了。谢谢你。

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
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2 回答 2

3

好的，那很快 :DI 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 count() in with ，然后按 with 排序，并收集DISTINCT作为回报。我不太确定这些带有语句的多个语句，会尝试清理它。;)

于 2014-08-06T22:22:09.260 回答
3 
MATCH (s:Store)-[:RELEASED]->(:Movie)<-[:ASSIGNED]-(cat:MovieCategorie)
WITH s, COUNT(cat) AS count, cat
ORDER BY s.name, count DESC
RETURN s.name AS Store, COLLECT(cat.name)[0..5] AS `Top Categories`

如果您想要viewsCount每个商店的 Movie 节点的属性总和：

MATCH (s:Store)-[:RELEASED]->(m:Movie)<-[:ASSIGNED]-(cat:MovieCategorie)
WITH s, COUNT(cat) AS count, m, cat
ORDER BY s.name, count DESC
RETURN s.name AS Store, COLLECT(cat.name)[0..5] AS `Top Categories`, SUM(m.viewsCount) AS `Total Views`
于 2014-08-06T22:24:10.787 回答