MYSQL 查询行组问题

Question

我正在尝试从我的 MySQL 数据库获取以下结果:

1) 我有一个包含以下 3 列的表格(time_in、time_out、receipt_number)

2)考虑在此表中我插入了以下行:

+--------------------+---------------------+----------------+
|     time_in        |     time_out        | receipt_number |
+--------------------+---------------------+----------------+
|2016-12-24 13:33:33 | 2016-12-24 13:52:52 |  1             |
|2016-12-24 13:44:44 | 2016-12-24 13:55:55 |  1             |
|2016-12-24 13:33:33 | 2016-12-24 13:55:55 |  2             |
|2016-12-24 13:35:35 | NULL                |  2             |
+--------------------+---------------------+----------------+

3)我需要从这些行中获取所有在“time_out”中包含相同receipt_number的行中具有NOT NULL值的收据(分组依据) >”列，这意味着我只需要提取收据编号 2。

SELECT DISTINCT receipt_number FROM orders 
WHERE time_in >= CURDATE() AND time_in < CURDATE() + INTERVAL 1 DAY 
(time_out IS NOT NULL OR time_out != "")
ORDER BY time_in ASC

但是，上述查询会返回两个收据编号，因为“receipt_number 2”包含一个非 NULL 值

我怎样才能实现这个目标？

谢谢

Answer 1

这是一种方法:

SELECT t1.*
FROM receipt_number t1
INNER JOIN
(
    SELECT receipt_number
    FROM orders
    GROUP BY receipt_number
    HAVING COUNT(*) = SUM(CASE WHEN time_out IS NOT NULL THEN 1 ELSE 0 END)
) t2
    ON t1.receipt_number = t2.receipt_number

别名为 t2 的子查询查找包含非 NULL 超时值的所有记录的收据编号。然后，它用于将 orders 表限制为您想要查看的记录。