SQL每个用户每天的第一个和最后一个时间戳

Question

我有一个userId和时间戳的表。对于每个用户，我想每天获得第一个和最后一个时间戳。我正在使用SQL Server。看起来如何：

|User|         Date        |
----------------------------
| 1  | 2015-01-08 07:31:40 |
| 1  | 2015-01-08 18:35:40 |
| 1  | 2015-01-08 22:31:40 |
| 1  | 2015-01-09 05:35:40 | 
| 1  | 2015-01-09 06:31:40 |
| 1  | 2015-01-09 07:35:40 | 
| 1  | 2015-01-09 12:38:40 | 
| 1  | 2015-01-09 23:38:40 |    
| 2  | 2015-01-08 05:55:40 |
| 2  | 2015-01-08 10:48:40 |
| 2  | 2015-01-08 15:38:40 |
| 2  | 2015-01-08 21:50:40 |
| 2  | 2015-01-09 10:38:40 |
| 2  | 2015-01-09 14:33:40 |
| 2  | 2015-01-09 15:38:40 |
| 2  | 2015-01-09 17:38:40 |
| 2  | 2015-01-09 19:55:40 |

我需要的输出：

|User|         Date        |
----------------------------
| 1  | 2015-01-08 07:31:40 |
| 1  | 2015-01-08 22:31:40 |
| 1  | 2015-01-09 05:35:40 |
| 1  | 2015-01-09 23:38:40 |
| 2  | 2015-01-08 05:55:40 |
| 2  | 2015-01-08 21:50:40 |
| 2  | 2015-01-09 10:38:40 |
| 2  | 2015-01-09 19:55:40 |

到目前为止它只适用于一个用户名：

SELECT 
    User, 
    Date, 
    min(Date) OVER(PARTITION BY Datepart(dy,Date)) as first, 
    max(Date) OVER(PARTITION BY Datepart(dy,Date)) as last
FROM 
    Tablename
WHERE 
    User = UserId

Answer 1

按用户和日期列分组。必须为其时间部分修剪日期列，如下所示：

SELECT MAX([Date]), MIN([Date]) FROM [TableName] GROUP BY [User], CONVERT(DATE, [Date])

语法对SQL Server有效，但方法应该与DBMS无关

Answer 2

SELECT 
    User, 
    Date, 
    min(Date) OVER(PARTITION BY Datepart(dy,Date)) as first, 
    max(Date) OVER(PARTITION BY Datepart(dy,Date)) as last
FROM 
    Tablename
GROUP BY
    Userid

这个Group-By声明可以提升您的代码，在Answer的评论部分添加您宝贵的反馈。