Question

我有这个表结构：

CREATE TABLE TIMETABLE
    ([ID] int, [Name] varchar(50), [StartDate] datetime, [EndDate] datetime)
;

INSERT INTO TIMETABLE
    ([ID], [Name], [StartDate], [EndDate])
VALUES
    (1, 'John',  '2017-01-29 16:00:00.000', '2017-01-29 16:12:00.000'),
    (2, 'Mario',  '2017-01-29 16:17:00.000', '2017-01-29 16:29:00.000'),
    (3, 'Kate',  '2017-01-15 10:35:00.000', '2017-01-15 10:40:00.000'),
    (4, 'Maria', '2017-01-15 10:17:00.000', '2017-01-15 10:27:00.000'),
    (5, 'Oliver',  '2017-01-15 13:46:00.000', '2017-01-29 14:00:00.000')
;

结果如下：

select * from TIMETABLE

ID  Name    StartDate               EndDate
1   John    2017-01-29T16:00:00Z    2017-01-29T16:12:00Z
2   Mario   2017-01-29T16:17:00Z    2017-01-29T16:29:00Z
3   Kate    2017-01-15T10:35:00Z    2017-01-15T10:40:00Z
4   Maria   2017-01-15T10:17:00Z    2017-01-15T10:27:00Z
5   Oliver  2017-01-15T13:46:00Z    2017-01-29T14:00:00Z

我想知道日期范围为15分钟的范围，例如：

   ID   Name    StartDate               EndDate                HourRangeTime
    1   John    2017-01-29T16:00:00Z    2017-01-29T16:12:00Z   16:00

在示例中，startdate和enddate在16:00到16:12之间，范围是16:00

结果应该是这样的：

ID  Name    StartDate               EndDate                   HourRangeTime
1   John    2017-01-29T16:00:00Z    2017-01-29T16:12:00Z      16:00
2   Mario   2017-01-29T16:17:00Z    2017-01-29T16:29:00Z      16:15
3   Kate    2017-01-15T10:35:00Z    2017-01-15T10:40:00Z      10:30
4   Maria   2017-01-15T10:17:00Z    2017-01-15T10:27:00Z      10:15
5   Oliver  2017-01-15T13:46:00Z    2017-01-29T14:00:00Z      13:45

如何填写HourRangeTime列，记录日期并查看它属于哪个范围？

Answer 1

您似乎专注于StartDate。

执行此操作的相对一般方法是将其转换为分钟，然后将分钟截断为最接近的15分钟。这是代码：

select cast(dateadd(minute,
                    15 * (datediff(minute, 0,
                                   cast(StartDate as time)
                                  ) / 15
                         ), 0
                   ) as time)

这会将结果作为time返回。

Answer 2

您可以获得差异并处理未来。

SELECT StartTime, EndTime, DATEDIFF(MINUTE, StartTime , EndTime) AS MinuteDiff 
FROM TIMETABLE

Answer 3

您可以尝试这个以获得所需的输出：

SELECT 
    CONCAT(DATEPART(hh,StartDate), ':', 
        CASE 
            WHEN DATEPART(MINUTE,StartDate) BETWEEN 0 AND 14 THEN '00'
            WHEN DATEPART(MINUTE,StartDate) BETWEEN 15 AND 29 THEN '15'
            WHEN DATEPART(MINUTE,StartDate) BETWEEN 30 AND 44 THEN '30'
            WHEN DATEPART(MINUTE,StartDate) BETWEEN 45 AND 59 THEN '45'
        ELSE '00'
        END) AS HourRangeTime 
FROM TIMETABLE

输出：

HourRangeTime
-------------
16:00
16:15
10:30
10:15
13:45

Answer 4

你可以使用它。

SELECT *,  
    CONVERT(VARCHAR,DATEPART(HOUR, [StartDate])) 
    + ':' 
    + RIGHT(CONVERT(VARCHAR,(DATEPART(MINUTE, [StartDate]) / 15) * 15)+'0',2) HourRangeTime FROM TIMETABLE

知道15分钟的日期间隔是SQL SERVER

4 个答案: