Question

我有这张桌子：

+----+-------+-------------------+
| ID | Value |       Date        |
+----+-------+-------------------+
| 1  | 8     | 12/10/2015 8:00am |
| 1  | 10    | 12/10/2015 8:30am |
| 1  | 5     | 12/10/2015 9:00am |
| 1  | 11    | 12/10/2015 9:30am |
| 2  | 8     | 12/10/2015 8:00am |
| 2  | 6     | 12/10/2015 8:30am |
| 2  | 8     | 12/10/2015 9:00am |
| 2  | 9     | 12/10/2015 9:30am |
+----+-------+-------------------+
etc.

现在我有一个这样的查询（为简洁起见而简化）：

SELECT ID, MAX(Value), MIN(Date)
FROM Values
WHERE ID IN (1,2,3,8)
GROUP BY (DATEPART(MINUTE, Date) / 60), ID
ORDER BY ID, MIN(Date)

返回的内容如下：

+----+-------+-------------------+
| ID | Value |       Date        |
+----+-------+-------------------+
| 1  | 10    | 12/10/2015 8:00am |
| 1  | 11    | 12/10/2015 9:00am |
| 2  | 8     | 12/10/2015 8:00am |
| 2  | 9     | 12/10/2015 9:00am |
| 3  | 12    | 12/10/2015 8:00am |
| 3  | 6     | 12/10/2015 9:00am |
| 8  | 51    | 12/10/2015 8:00am |
| 8  | 58    | 12/10/2015 9:00am |
+----+-------+-------------------+

我需要计算值字段的总和，如果，比方说，ID是1,2或3 我期待这样的事情：

+------------+-------+-------------------+
|  IDfield   | Value |       Date        |
+------------+-------+-------------------+
| IDstring1  | 30    | 12/10/2015 8:00am |
| IDstring1  | 26    | 12/10/2015 9:00am |
| IDstring2  | 51    | 12/10/2015 8:00am |
| IDstring2  | 58    | 12/10/2015 9:00am |
+------------+-------+-------------------+

或者，甚至更好：

+--------+--------+-------------------+
| Value1 | Value2 |       Date        |
+--------+--------+-------------------+
| 30     | 51     | 12/10/2015 8:00am |
| 26     | 58     | 12/10/2015 9:00am |
+--------+--------+-------------------+

我需要求和的ID是静态的，所以我想我可以选择其中一种解决方案，但我想不出办法这样做。

我会尝试更好地解释预期结果：

Value1 应包含值字段的SUM，其中ID为IN（1,2,3），按小时分组，如我在查询中所做的那样 Value2 应该只包含ID字段，其中ID = 8

编辑：添加了源表示例，更正了当前查询（我忘记将聚合MAX（）放入SELECT中的字段值

编辑＃2：添加了更好的解释

Answer 1

数据库：PostgreSql

查询：

select sum("Value") as Value1, date from "Values" where id in(1,2,3) group by date
Union ALL
select sum("Value") as Value1, date from "Values" where id=8 group by date

Answer 2

我想我终于得到了你想要的东西：

DECLARE @tbl TABLE(ID INT, Value INT, Date DATETIME);
INSERT INTO @tbl VALUES
 (1,10,'12/10/2015 8:00am')
,(1,11,'12/10/2015 9:00am')
,(2,8,'12/10/2015 8:00am')
,(2,9,'12/10/2015 9:00am')
,(3,12,'12/10/2015 8:00am')
,(3,6,'12/10/2015 9:00am')
,(8,51,'12/10/2015 8:00am')
,(8,58,'12/10/2015 9:00am');

WITH Values123 AS
(
    SELECT SUM(Value) AS Sum123,Date
    FROM @tbl
    WHERE ID IN(1,2,3)
    GROUP BY Date
)
SELECT Values123.Sum123
      ,(SELECT SUM(Value) FROM @tbl WHERE Date=Values123.Date AND ID=8)
      ,Values123.Date 
FROM Values123

结果

Sum123  Sum8    Date
30      51      2015-10-12 08:00:00.000
26      58      2015-10-12 09:00:00.000

如果其ID在定义的范围内，则对行进行求值

2 个答案: