通过以下方式检索组中的多个信息

Question

我有一个包含 3 列 Serviceid、Date、User_id 的表“购买”

一些数据如下所示：

20  2-Jan-18    40709217
20  2-Jan-18    40709217
40  2-Jan-18    40709217
40  2-Jan-20    40709217
50  2-Jan-21    40709217    
984 22-Mar-18   18246539    
269 22-Mar-18   18246539    
666 1-Apr-18    18246539

我的查询请求是：

对于每个 'user_id'，获取以下信息：

• 用户购买的前 2 个最早的 ServiceId 和日期
• 用户最近购买的 ServiceId 和 Date
• 用户购买的服务计数

结果表的列必须遵循以下顺序： User_id、FirstServiceid、SecondServiceid、FirstServiceDate、SecondServiceDate、LastServiceid、LastServiceDate、TotalService。

预期输出：

User_id	FirstServiceid	SecondServiceid	FirstDate	SecondDate	LastServiceid	LastDate	TotalServices
40709217	20	40	2-Jan-18	2-Jan-20	50	2-Jan-21	5
18246539	984	666	22-Mar-18	1-Apr-18	666	1-Apr-18	3

我的想法是进行聚合，然后将它们连接在一起，但我遇到了这个错误

“列 'purchase.Serviceid' 在选择列表中无效，因为它 不包含在聚合函数或 GROUP BY 中 条款”

调用最新的Serviceid和Date时：

select user_id, max(date), serviceid    
from purchases    
group by user_id

如何克服这个问题，有没有更好的方法来聚合大量信息而无需使用 JOIN？

Answer 1

您可以尝试使用以下使用窗口函数（例如 row_number 和 dense_rank）在通过 user_id 聚合结果之前识别所需的记录，并在 MAX 函数中使用 case 表达式来提取数据所需的列

SELECT
    User_id,
    MAX(CASE WHEN earliest_order=1 and rp=1 THEN Serviceid END) as FirstServiceid,
    MAX(CASE WHEN earliest_order=2 and rp=1 THEN Serviceid END) as SecondServiceid,
    MAX(CASE WHEN earliest_order=1 and rp=1 THEN [Date] END) as FirstServiceDate,
    MAX(CASE WHEN earliest_order=2 and rp=1 THEN [Date] END) as SecondServiceDate,
    MAX(CASE WHEN latest_order=1 THEN Serviceid END) as LastServiceid,
    MAX(CASE WHEN latest_order=1 THEN [Date] END) as LastServiceDate,
    COUNT(1) as TotalService
FROM (
   SELECT 
       *,
       DENSE_RANK() OVER (
           PARTITION BY User_id
           ORDER BY [Date] 
       ) as earliest_order,
       ROW_NUMBER() OVER (
           PARTITION BY User_id
           ORDER BY [Date] DESC, Serviceid
       ) as latest_order,
       ROW_NUMBER() OVER (
              PARTITION BY User_id,[Date]
              ORDER BY Serviceid DESC
       ) rp
   FROM 
       purchases
   
) t
GROUP BY
    User_id
ORDER BY
    User_id DESC

上面使用的子查询的结果也包含在下面，供您阅读。

   SELECT
        *,
       DENSE_RANK() OVER (
           PARTITION BY User_id
           ORDER BY [Date] 
       ) as earliest_order,
       ROW_NUMBER() OVER (
           PARTITION BY User_id
           ORDER BY [Date] DESC, Serviceid
       ) as latest_order,
       ROW_NUMBER() OVER (
              PARTITION BY User_id,[Date]
              ORDER BY Serviceid DESC
       ) rank_priority
   FROM
           purchases
    ORDER BY User_id, [Date], Serviceid DESC

Serviceid	Date	User_id	earliest_order	latest_order	rank_priority
984	2018-03-22	18246539	1	3	1
269	2018-03-22	18246539	1	2	2
666	2018-04-01	18246539	2	1	1
40	2018-01-02	40709217	1	5	1
20	2018-01-02	40709217	1	3	2
20	2018-01-02	40709217	1	4	3
40	2020-01-02	40709217	2	2	1
50	2021-01-02	40709217	3	1	1

View working demo db fiddle here