Question

数据如下：

date_time        | amount| user|group
---------------------
5/15/2017 6:30 | 250| a | abc
5/15/2017 6:30 | 250| a | def
5/15/2017 5:30 | 50| a |abc
5/15/2017 5:30 | 50| a |def
5/15/2017 4:30 | 50| a |abc
5/15/2017 4:30 | 50| a |def
5/15/2017 5:20 | 50| b |abc
5/15/2017 5:20 | 50| b |def
5/15/2017 5:18 | 50| b |abc
5/15/2017 5:18 | 50| b |def

期望的结果：

date_time| amount | user  | group
---------------------
5/15/2017 4:30 | 50| a |abc
5/15/2017 4:30 | 50| a |def
5/15/2017 6:30 | 250| a | abc
5/15/2017 6:30 | 250| a | def
5/15/2017 5:18 | 50| b |abc
5/15/2017 5:18 | 50| b |def

具有不同值而不考虑日期时间列的结果，对于整个组，不应该对日期时间应用不同。

Answer 1

您可以按行，用户，组使用row_number和分区，如下所示：

select top 1 with ties * from YourdateTable
    order by row_number() over (partition by amount, [user], [group] order by amount, [user], [group])

输出：

+-------------------------+--------+------+-------+
|        date_time        | amount | user | group |
+-------------------------+--------+------+-------+
| 2017-05-15 05:30:00.000 |     50 | a    | abc   |
| 2017-05-15 04:30:00.000 |     50 | a    | def   |
| 2017-05-15 05:20:00.000 |     50 | b    | abc   |
| 2017-05-15 05:18:00.000 |     50 | b    | def   |
| 2017-05-15 06:30:00.000 |    250 | a    | abc   |
| 2017-05-15 06:30:00.000 |    250 | a    | def   |
+-------------------------+--------+------+-------+

Answer 2

;With cte(date_time, amount,[user],[group])
AS
(
SELECT '5/15/2017 6:30' , 250, 'a' , 'abc' Union all
SELECT '5/15/2017 6:30' , 250, 'a' ,'def'  Union all
SELECT '5/15/2017 5:30' , 50, 'a' ,'abc'   Union all
SELECT '5/15/2017 5:30' , 50, 'a' ,'def'   Union all
SELECT '5/15/2017 4:30' , 50, 'a' ,'abc'   Union all
SELECT '5/15/2017 4:30' , 50, 'a' ,'def'   Union all
SELECT '5/15/2017 5:20' , 50, 'b' ,'abc'   Union all
SELECT '5/15/2017 5:20' , 50, 'b' ,'def'   Union all
SELECT '5/15/2017 5:18' , 50, 'b' ,'abc'   Union all
SELECT '5/15/2017 5:18' , 50, 'b' ,'def'
)

SELECT date_time
    ,amount
    ,[user]
    ,[group]
FROM (
    SELECT *
        ,ROW_NUMBER() OVER (
            PARTITION BY amount
            ,[user]
            ,[group] ORDER BY amount
                ,[user]
            ) seq
    FROM cte
    ) dt
WHERE dt.seq = 1
ORDER BY [user]

以下是输出：

date_time   amount  user    group
--------------------------------
5/15/2017   4:30    50  a   abc
5/15/2017   5:30    50  a   def
5/15/2017   6:30    250 a   abc
5/15/2017   6:30    250 a   def
5/15/2017   5:18    50  b   abc
5/15/2017   5:18    50  b   def

Answer 3

SELECT MIN(date_time) AS date_time, amount, [user], [group]
FROM MyTable
GROUP BY [user], amount, [group];

根据您的上述数据，您选择的日期无关紧要，因此 MIN（）用于选择不在 GROUP BY 中的日期，即不明显。< / p>

获取除一个列值

3 个答案: