Question

我有一个看起来像这样的表：

code int, primary key
reservation_code int,
indate date,
outdate date,
slot int,
num int,

数据库有一些奇怪的设计，它应该工作的方式是这个表保存每个槽预订的日期，num用于跟踪我认为是遗留原因的连续预订。 / p>

我正在尝试提出一个检查数据库中先前预订的查询。我这样做的想法：

对于给定的插槽号，检查是否存在一组具有相同reservation_code的行，该行具有该组的最小num值的行的日期小于或等于给定日期和过期日期在具有最大num值的行上高于相同的给定日期。

我在SQL中最接近这个：

编辑：在Barmar的帮助下

SELECT b.reservation_code
FROM bookings b
JOIN (SELECT reservation_code, MIN(num) minnum
      FROM bookings
      WHERE slot = "given_slot"
      AND indate <= "given_date"
      GROUP BY reservation_code) min
ON minnum = num and b.reservation_code = min.reservation_code
JOIN (SELECT reservation_code, MAX(num) maxnum
      FROM bookings
      WHERE slot = "given_slot"
      AND outdate > "given_date"
      GROUP BY reservation_code) max
ON maxnum = num and b.reservation_code = max.reservation_code
WHERE slot="given_slot"
AND indate <= "given_date"
AND outdate > "given_date"
GROUP BY b.reservation_code

将GROUP BY添加到两个子查询中使其适用于大多数情况，但第二次检查仍会返回错误的答案。

以下是一些示例行和查询，试图让问题更加清晰：

示例行：

code    reservation_code    indate      outdate     slot    num
1       1                   01/01/13    03/01/13    1       0
2       1                   03/01/13    05/01/13    1       1
3       1                   05/01/13    10/01/13    1       2
4       2                   04/01/13    15/01/13    2       0
5       2                   15/01/13    19/01/13    2       1
6       3                   11/01/13    13/01/13    1       0
7       4                   15/01/13    16/01/13    3       0
8       5                   01/01/13    15/01/13    3       0
9       5                   15/01/13    25/01/13    4       1

抽样检查：

slot 2, date 21/02/13, should return not booked.
slot 2, date 16/01/13, should return booked
slot 1, date 14/01/13, should return not booked
slot 1, date 12/01/13, should return booked
slot 1, date 10/01/13, should return not booked
slot 3, date 02/01/13, should return booked
slot 4, date 15/01/13, should return booked
slot 4, date 25/01/13, should return not booked

Answer 1

您需要对聚合表

使用JOIN

SELECT b.reservation_code, count(1)
FROM bookings b
JOIN (SELECT reservation_code, MAX(num) maxnum
      FROM bookings
      WHERE slot = "given slot"
      AND indate <= "given date"
      GROUP BY reservation_code) m
ON maxnum = num and b.reservation_code = m.reservation_code
WHERE slot="given slot"
AND indate <= "given date"
GROUP BY b.reservation_code

Answer 2

经过一夜好眠，我意识到我的问题相当简单，可以通过一个非常简单的查询解决，例如：

SELECT 1
FROM bookings
WHERE slot="given slot"
AND indate <= "given date"
AND outdate > "given date"

我要感谢所有试图帮助我的人，对不起，我浪费了你的时间。

SQL获取比较聚合函数的结果

2 个答案: