t_table
好像:
+-----------+---------+--------------+------------------+-----------------------+----------------------------------+
| pk_IdLoan | fk_IdCar| fk_IdCustomer| fk_Source_Agency | fk_Destination_Agency | RentalDate | DeliveryDate | Cost |
+-----------+---------+--------------+------------------+-----------------------+----------------------------------+
我写了一个查询:
(SELECT fk_IdCustomer, MONTHNAME(RentalDate) AS Month, YEAR(RentalDate) As Year, COUNT(*)
FROM t_loan
GROUP BY fk_IdCustomer, Month, Year);
这导致
+---------------+-------------+------+----------+
| fk_IdCustomer | Month | Year | COUNT(*) |
+---------------+-------------+------+----------+
| 1 | July | 2016 | 3 |
| 1 | November | 2017 | 1 |
| 1 | September | 2016 | 7 |
| 5 | May | 2016 | 1 |
| 6 | January | 2016 | 1 |
| 6 | September | 2017 | 2 |
+---------------+-------------+------+----------+
现在我想为每个客户获得这些月份和年份,从而获得最高的COUNT(*)
fe:
+---------------+-------------+------+----------+
| fk_IdCustomer | Month | Year | COUNT(*) |
+---------------+-------------+------+----------+
| 1 | September | 2016 | 7 |
| 5 | May | 2016 | 1 |
| 6 | September | 2017 | 2 |
+---------------+-------------+------+----------+
如何实现这一目标?
这在 MySQL 中有点痛苦,它不支持 CTE 或窗口函数。一种方法是:
SELECT fk_IdCustomer, MONTHNAME(RentalDate) AS Month,
YEAR(RentalDate) As Year, COUNT(*) as cnt
FROM t_loan l
GROUP BY fk_IdCustomer, Month, Year
HAVING cnt = (SELECT COUNT(*)
FROM t_loan l2
WHERE l2.fk_IdCustomer = l.fk_IdCustomer
GROUP BY MONTHNAME(RentalDate), YEAR(RentalDate)
ORDER BY COUNT(*) DESC
LIMIT 1
);
注意:如果有重复,您将获得所有匹配的值。
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