返回具有复杂 where 条件的查询

Question

我有一张这样的桌子

表allotment

allotment_id | date | room_id | is_closed
-----------------------------------------
1 | 2017-01-28 | 21 | 0
2 | 2017-01-29 | 21 | 1
3 | 2017-01-30 | 21 | 0
4 | 2017-01-31 | 21 | 0
5 | 2017-01-28 | 32 | 0
6 | 2017-01-29 | 32 | 0
7 | 2017-01-30 | 32 | 0
8 | 2017-01-31 | 32 | 0

我正在尝试这样的查询：

SELECT *
FROM allotment
WHERE date BETWEEN '2017-01-28' AND DATE_SUB('2017-01-31', INTERVAL 1 DAY)
AND is_closed = 0
AND date >= '2017-01-28' AND l.`date` <= DATE_SUB('2017-01-31', INTERVAL 1 DAY) AND allotment_id NOT IN (select allotment_id FROM allotment WHERE is_closed = 1 AND date >= '2017-01-28' AND date <= '2017-01-31')

如您所见，我选择日期为 2017-01-28 到 2017-01-31，桌子上有两个房间 ID，room_id 21 日期为 2017-01-29 is_closed 为 1。

我想要的只是返回 room_id 32，因为在 room_id 21 中，我选择的日期范围是 is_closed 中的值 1。 p>

我已经尝试过我的查询，但它不起作用。我怎么能这样做。

谢谢。

Answer 1

在哪里和子查询中，您使用了 allotment_id。而不是这个，你必须带房间号

SELECT *
FROM allotment
WHERE date BETWEEN '2017-01-28' AND DATE_SUB('2017-01-31', INTERVAL 1 DAY)
AND is_closed = 0 and room_id not in (select room_id FROM allotment WHERE is_closed = 1 AND date >= '2017-01-28' AND date <= '2017-01-31')

Answer 2

按房间分组，只选择那些is_closed 标志为零次的人

SELECT room_id
FROM allotment
WHERE date >= '2017-01-28' 
  AND date <= '2017-01-31'
group by room_id
having sum(is_closed = 1) = 0