对另一个查询的结果执行查询？

Posted 2023-02-16

【中文标题】对另一个查询的结果执行查询？

【英文标题】：Performing a query on a result from another query?

【发布时间】：2010-10-31 06:31:19

【问题描述】：

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

返回类似：

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？只返回一行包含 Count 和 SUM？

Count         SUM
5             3

谢谢

【参考方案1】：

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;

【参考方案2】：

通常您可以插入查询的结果（基本上是一个表）作为 FROM 子句源 另一个查询，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

【讨论】：

问题，不会为外部查询的每条记录运行子查询吗？或者结果会被缓存

sql server 最有可能做的就是运行一次子查询，然后对结果进行流聚合。您可以使用Include Actual Execution Plan 选项查看它的实际作用。

【参考方案3】：

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。

【参考方案4】：

请注意，您的初始查询可能没有返回您想要的：

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