具有复杂查询的预制列上的 sql sum 或聚合函数答案

【问题标题】：sql sum or aggregate function on pre-fab column with complex query具有复杂查询的预制列上的 sql sum 或聚合函数
【发布时间】：2009-12-02 12:15:32
【问题描述】：

我有一个查询，它从一周中的 7 天长时间获取平均数据（价格）。周一、周二等的 IE 平均价格。它工作正常，但我不确定如何在同一个查询中求和该查询找到的平均价格？总结 Day1..Day5

就目前而言，此查询将所有价格的全部总和... IE 巨大的数字.. 不是来自平均值。

有什么想法吗？顺便说一句，这是来自 Postgres...谢谢

SELECT rooms.name, rooms.id,
MAX(CASE WHEN (EXTRACT(DOW FROM availables.bookdate) - EXTRACT(DOW FROM DATE '2009-12-13')) = 0 THEN (availables.price) ELSE 0 END) AS day1,
MAX(CASE WHEN (EXTRACT(DOW FROM availables.bookdate) - EXTRACT(DOW FROM DATE '2009-12-13')) = 1 THEN (availables.price) ELSE 0 END) AS day2,
MAX(CASE WHEN (EXTRACT(DOW FROM availables.bookdate) - EXTRACT(DOW FROM DATE '2009-12-13')) = 2 THEN (availables.price) ELSE 0 END) AS day3,
MAX(CASE WHEN (EXTRACT(DOW FROM availables.bookdate) - EXTRACT(DOW FROM DATE '2009-12-13')) = 3 THEN (availables.price) ELSE 0 END) AS day4,
MAX(CASE WHEN (EXTRACT(DOW FROM availables.bookdate) - EXTRACT(DOW FROM DATE '2009-12-13')) = 4 THEN (availables.price) ELSE 0 END) AS day5,
(AVG(availables.price)) AS avg,
(SUM(availables.price)) AS sum, MAX((SIGN(spots)-1) + 1) AS beds
 FROM availables
 INNER JOIN rooms
 ON availables.room_id=rooms.id
 WHERE availables.room_id = '1780' AND availables.price > 0
 GROUP BY rooms.id, rooms.name

【问题讨论】：

标签： sql postgresql aggregate sum average

【解决方案1】：

这是应该起作用的东西....

select avg(mon+tues+weds+thur+fri) as averageall,
       sum(mon+tues+weds+thur+fri) as sumall,
  avg(mon) as avgmon,
  avg(tues) as avgtues,
  avg(weds) as avgweds,
  avg(thurs) as avgthurs,
  agv(fri) as avgfri,
MAX((SIGN(spots)-1) + 1) AS beds
from
(
SELECT rooms.name, rooms.id,
CASE WHEN EXTRACT(DOW FROM availables.bookdate) = 1 THEN (availables.price) ELSE 0 END) AS mon,
CASE WHEN EXTRACT(DOW FROM availables.bookdate) = 2 THEN (availables.price) ELSE 0 END) AS tues,
CASE WHEN EXTRACT(DOW FROM availables.bookdate) = 3 THEN (availables.price) ELSE 0 END) AS weds,
CASE WHEN EXTRACT(DOW FROM availables.bookdate) = 4 THEN (availables.price) ELSE 0 END) AS thur,
CASE WHEN EXTRACT(DOW FROM availables.bookdate) = 5 THEN (availables.price) ELSE 0 END) AS fri,
spots, rooms.id, rooms.name
 FROM availables
 INNER JOIN rooms
 ON availables.room_id=rooms.id
 WHERE availables.room_id = '1780' AND availables.price > 0
) 
 GROUP BY rooms.id, rooms.name
)

注意我没有测试所以可能有错别字。

【讨论】：