37

我有一个查询：

SELECT availables.bookdate AS Date, DATEDIFF(now(),availables.updated_at) as Age
FROM availables
INNER JOIN rooms
ON availables.room_id=rooms.id
WHERE availables.bookdate BETWEEN '2009-06-25' AND date_add('2009-06-25', INTERVAL 4 DAY) AND rooms.hostel_id = 5094
GROUP BY availables.bookdate

它返回如下内容：

Date               Age
2009-06-25         0
2009-06-26         2
2009-06-27         1
2009-06-28         0
2009-06-29         0

然后我怎样才能对返回的行数进行计数..（在本例中为 5）和年龄的总和？只返回一行的计数和总和？

Count         SUM
5             3

谢谢

4

4 回答 4

68

通常你可以插入一个查询的结果（基本上是一个表）作为另一个查询的 FROM 子句源，所以会这样写：

SELECT COUNT(*), SUM(SUBQUERY.AGE) from
(
  SELECT availables.bookdate AS Date, DATEDIFF(now(),availables.updated_at) as Age
  FROM availables
  INNER JOIN rooms
  ON availables.room_id=rooms.id
  WHERE availables.bookdate BETWEEN '2009-06-25' AND date_add('2009-06-25', INTERVAL 4 DAY) AND rooms.hostel_id = 5094
  GROUP BY availables.bookdate
) AS SUBQUERY
于 2009-06-04T09:30:15.470 回答
14

您只需将查询包装在另一个中：

SELECT COUNT(*), SUM(Age)
FROM (
    SELECT availables.bookdate AS Count, DATEDIFF(now(),availables.updated_at) as Age
    FROM availables
    INNER JOIN rooms
    ON availables.room_id=rooms.id
    WHERE availables.bookdate BETWEEN '2009-06-25' AND date_add('2009-06-25', INTERVAL 4 DAY) AND rooms.hostel_id = 5094
    GROUP BY availables.bookdate
) AS tmp;
于 2009-06-04T09:27:50.203 回答
2

我不知道你是否甚至需要包装它。这行不通吗？

SELECT COUNT(*), SUM(DATEDIFF(now(),availables.updated_at))
FROM availables
INNER JOIN rooms    ON availables.room_id=rooms.id
WHERE availables.bookdate BETWEEN '2009-06-25' 
  AND date_add('2009-06-25', INTERVAL 4 DAY)
  AND rooms.hostel_id = 5094
GROUP BY availables.bookdate);

如果您的目标是返回两个结果集，那么您需要将其临时存储在某个地方。

于 2009-06-04T09:32:08.167 回答
0

请注意，您的初始查询可能不会返回您想要的内容：

SELECT availables.bookdate AS Date, DATEDIFF(now(),availables.updated_at) as Age 
FROM availables INNER JOIN rooms ON availables.room_id=rooms.id 
WHERE availables.bookdate BETWEEN  '2009-06-25' AND date_add('2009-06-25', INTERVAL 4 DAY) AND rooms.hostel_id = 5094 GROUP BY availables.bookdate

您正在按预订日期分组，但您没有在查询的第二列上使用任何分组功能。

您可能正在寻找的查询是：

SELECT availables.bookdate AS Date, count(*) as subtotal, sum(DATEDIFF(now(),availables.updated_at) as Age)
FROM availables INNER JOIN rooms ON availables.room_id=rooms.id
WHERE availables.bookdate BETWEEN '2009-06-25' AND date_add('2009-06-25', INTERVAL 4 DAY) AND rooms.hostel_id = 5094
GROUP BY availables.bookdate
于 2009-06-04T10:06:54.727 回答