为 MYSQL 中某个时间段内的每一天添加行

Question

我正在处理“预订”表 -

我想在此表中添加“date_of_stay”列，其中 date_of_stay 将根据“nights”列给出的晚数存储 booking_id 停留期间的每个日期。 例如-

booking_id    booking_date        nights    date_of_stay

5001        Thu, 03 Nov 2016       7        Thu, 03 Nov 2016
5001        Thu, 03 Nov 2016       7        Fri, 04 Nov 2016
5001        Thu, 03 Nov 2016       7        Sat, 05 Nov 2016
5001        Thu, 03 Nov 2016       7        Sun, 06 Nov 2016
5001        Thu, 03 Nov 2016       7        Mon, 07 Nov 2016
5001        Thu, 03 Nov 2016       7        Tue, 08 Nov 2016
5001        Thu, 03 Nov 2016       7        Wed, 09 Nov 2016
5002        Thu, 03 Nov 2016       2        Thu, 03 Nov 2016
5002        Thu, 03 Nov 2016       2        Fri, 04 Nov 2016

在不改变表格的情况下，以这种方式查看我的表格的最简单方法是什么？

Answer 1

使用递归 CTE：

with cte as (
      select booking_id, booking_date, nights, 1 as n
      from t
      union all
      select booking_id, booking_date, nights, 1 + n
      from cte
      where n < nights
     )
select *
from cte;

Answer 2

根据article，您可以使用下一个查询：

select booking.*, date_of_stay
from (
  select adddate('2015-01-01', t3*1000 + t2*100 + t1*10 + t0) date_of_stay from
    (select 0 t0 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t0,
    (select 0 t1 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t1,
    (select 0 t2 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t2,
    (select 0 t3 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t3
) v
join booking ON date_of_stay between 
    booking.booking_date and date_add(booking.booking_date, interval nights-1 day)
order by booking_id, date_of_stay
;

上述查询也适用于 MYSQL 5 和 8。对区间内的所有日期都有效：

select '2015-01-01' from_date, adddate('2015-01-01',9*1000 + 9*100 + 9*10 + 9) to_date;
+------------+------------+
|  from_date |    to_date |
+------------+------------+
| 2015-01-01 | 2042-05-18 |
+------------+------------+

这里查询可以测试SQLize.online

Answer 3

您可以使用递归 CTE 来做到这一点：

WITH RECURSIVE cte AS (
  SELECT booking_id, booking_date, nights, 0 nr, 
         STR_TO_DATE(booking_date, '%a, %d %b %Y') date_of_stay
  FROM booking
  UNION ALL
  SELECT booking_id, booking_date, nights, nr + 1,
         date_of_stay + interval 1 day
  FROM cte       
  WHERE nr < nights - 1       
)
SELECT c.booking_id, c.booking_date, c.nights, 
       DATE_FORMAT(c.date_of_stay, '%a, %d %b %Y') date_of_stay
FROM cte c
ORDER BY c.booking_id, c.date_of_stay

请参阅demo。
结果：

> booking_id | booking_date     | nights | date_of_stay    
> ---------: | :--------------- | -----: | :---------------
>       5001 | Thu, 03 Nov 2016 |      7 | Thu, 03 Nov 2016
>       5001 | Thu, 03 Nov 2016 |      7 | Fri, 04 Nov 2016
>       5001 | Thu, 03 Nov 2016 |      7 | Sat, 05 Nov 2016
>       5001 | Thu, 03 Nov 2016 |      7 | Sun, 06 Nov 2016
>       5001 | Thu, 03 Nov 2016 |      7 | Mon, 07 Nov 2016
>       5001 | Thu, 03 Nov 2016 |      7 | Tue, 08 Nov 2016
>       5001 | Thu, 03 Nov 2016 |      7 | Wed, 09 Nov 2016
>       5002 | Thu, 03 Nov 2016 |      2 | Thu, 03 Nov 2016
>       5002 | Thu, 03 Nov 2016 |      2 | Fri, 04 Nov 2016