mysql - SQL简化一个特征函数？或 PIVOT 表？

Question

我有一张包含一堆日期和价格的表格：

房间名称、价格、预订日期等

我可以像这样转换它：（基本上翻转列）

SELECT availables.name, rooms.id,
  MAX(IF(to_days(availables.bookdate) - to_days('2009-06-13') = 0, availables.price, '')) AS day1,
  MAX(IF(to_days(availables.bookdate) - to_days('2009-06-13') = 1, availables.price, '')) AS day2,
  MAX(IF(to_days(availables.bookdate) - to_days('2009-06-13') = 2, availables.price, '')) AS day3,
  MAX(IF(to_days(availables.bookdate) - to_days('2009-06-13') = 3, availables.price, '')) AS day4,
  MAX(IF(to_days(availables.bookdate) - to_days('2009-06-13') = 4, availables.price, '')) AS day5,
  MAX(IF(to_days(availables.bookdate) - to_days('2009-06-13') = 5, availables.price, '')) AS day6,
  MAX(IF(to_days(availables.bookdate) - to_days('2009-06-13') = 6, availables.price, '')) AS day7,
AVG(availables.price),SUM(availables.price)
FROM `availables`
INNER JOIN rooms
ON availables.room_id=rooms.id
WHERE availables.room_id = '18382'
GROUP BY availables.name

这完美地工作并产生了这个：

name    id  day1    day2    day3    day4    day5    day6    day7    AVG(availables.price)   SUM(availables.price)
Bed     18382   23.00   21.00   21.00   21.00   21.00   21.00       21.571429   151.00

但是由于我不知道天数，我该如何简化呢？它可能是1或7？有任何想法吗？

Answer 1

看看这个答案及其链接的答案。