3 每个客户的最后记录答案

【问题标题】：3 last record of each customer3 每个客户的最后记录
【发布时间】：2014-07-31 16:21:46
【问题描述】：

我有一张这样的客户和请求表：

客户表：

Key | Name
----+-----------
  1 | Roberto
  2 | Thiago
  3 | Mike

请求表：

key | Date       | Customer
----+------------+------------  
  1 | 2012-02-07 | 1   
  2 | 2012-02-08 | 2
  3 | 2012-02-09 | 1
  4 | 2012-03-07 | 1
  5 | 2012-03-08 | 3
  6 | 2012-03-09 | 2
  7 | 2012-04-07 | 3
  8 | 2012-04-08 | 1
  9 | 2012-04-09 | 3

我想要一个返回每个客户的最后 3 个请求的查询。 Obs：我正在使用 MySQL 服务器

返回应该是这样的：

key | Date       | Customer
----+------------+-----------
  1 | 2012-02-07 | 1
  3 | 2012-02-09 | 1
  4 | 2012-03-07 | 1
  2 | 2012-02-08 | 2
  6 | 2012-03-09 | 2
  5 | 2012-03-08 | 3
  7 | 2012-04-07 | 3
  9 | 2012-04-09 | 3

我不能使用命令“TOP”，因为我使用的是 MySQL 服务器，而这个命令只能在 SQL Server 中使用。

【问题讨论】：

标签： mysql sql greatest-n-per-group

【解决方案1】：

你可以试试

select 
r.`key`,
r.Date,
r.customer
from Customers c
left join 
(
  select 
  r1.*
  from Requests r1
  where 
  (
    select count(*)
    from Requests r2
    where r1.customer = r2.customer
    AND r1.`key` <= r2.`key`
  ) <=3
  order by r1.Date desc
)r
on r.customer = c.`key`
order by c.`key`

另一种方式

select
r.`key`,
r.Date,
r.customer
from Customers c
join Requests r on r.`Customer` = c.`key`
where
(
  select count(*) from Customers c1
  join Requests r1 on r1.`Customer` = c1.`key`
  where
  c.`key` = c1.`key`
  and r1.`key`>= r.`key`
) <=3
order by c.`key`,r.Date desc

DEMO

【讨论】：

【解决方案2】：

您可以查看此here。

SELECT DISTINCT `key`, date, customer
FROM
(
        SELECT  MAX(r1.`key`) `key`, MAX(r1.date) date, r1.customer customer
        FROM requests r1
        GROUP BY r1.customer
    UNION
        SELECT MAX(r2.`key`) `key`, MAX(r2.date) date, r1.customer customer
        FROM requests r1
        JOIN requests r2 ON  r2.customer = r1.customer
                              AND r2.date < r1.date
        GROUP BY r1.customer
    UNION
        SELECT MAX(r3.`key`) `key`, MAX(r3.date) date, r1.customer customer
        FROM requests r1
        JOIN requests r2 ON  r2.customer = r1.customer
                              AND r2.date < r1.date
        JOIN requests r3 ON r3.customer = r2.customer
                              AND r3.date < r2.date
        GROUP BY r1.customer
) subquery
ORDER BY customer, date, `key`

【讨论】：

我误解了这个问题。我用“last”来表示“最新”或“最近”的记录。无论如何，我更喜欢@Abhik-Chakraborty 的回答！

【解决方案3】：

我想这应该可行：

SELECT     req.Key,
           req.date,
           req.customer
FROM       [dbo].[Requests] req
INNER JOIN [dbo].[Customers] cust ON cust.CustomerID = req.customerID
GROUP BY   req.Key, req.date, req.customer
ORDER BY   req.date

希望对你有帮助！！！

【讨论】：

感谢 Satwik，但我使用的是 MySQL Server，无法使用 TOP，只能在 SQL Server 中使用。另一个建议？