Question

我在sql server 2012中有一个表，其中列名为Duration as Time datatype.i想要根据日期差异更新此列，方法是将差异添加到此持续时间列中。我可以在SP中执行此操作。 / p>

ID    StartDate               EndDate                 Duration
1     2017-02-27 09:10:35     2017-02-27 09:25:35      00:15
2     2017-02-27 09:26:35   2017-02-27 09:36:35        00:25

Durtion总是不到24小时。

Answer 1

一种方法是：

update t
    set duration = cast(dateadd(ms, datediff(ms, startdate, enddate), 0) as time);

Answer 2

这很简单。只需使用简单的算术运算：

DECLARE @TABLE TABLE (Start DATETIME, Finish DATETIME, Duration TIME);

INSERT INTO @TABLE VALUES ('02/28/2017 08:00','02/28/2017 08:30','');
INSERT INTO @TABLE VALUES ('02/28/2017 09:00','02/28/2017 09:40','');
INSERT INTO @TABLE VALUES ('02/28/2017 10:02','02/28/2017 11:53','');
INSERT INTO @TABLE VALUES ('02/28/2017 11:56','02/28/2017 12:45','');
INSERT INTO @TABLE VALUES ('02/28/2017 13:45','02/28/2017 23:59','');

UPDATE @TABLE
SET Duration = Finish - Start;

SELECT * FROM @TABLE;

返回：

Start                    Finish                 Duration
---------------------------------------------------------
28/02/2017 08:00:00      28/02/2017 08:30:00    00:30:00
28/02/2017 09:00:00      28/02/2017 09:40:00    00:40:00
28/02/2017 10:02:00      28/02/2017 11:53:00    01:51:00
28/02/2017 11:56:00      28/02/2017 12:45:00    00:49:00
28/02/2017 13:45:00      28/02/2017 23:59:00    10:14:00

唯一需要注意的是，他们需要在同一天。你明确表示持续时间不会超过一天，所以应该没问题。

如果您想将结果添加到Duration的原始值，那么您只需将其添加到...

INSERT INTO @TABLE VALUES ('02/27/2017 08:00','02/28/2017 08:30','00:15');
INSERT INTO @TABLE VALUES ('02/28/2017 09:00','02/28/2017 09:40','00:14');
INSERT INTO @TABLE VALUES ('02/28/2017 10:02','02/28/2017 11:53','00:13');
INSERT INTO @TABLE VALUES ('02/28/2017 11:56','02/28/2017 12:45','02:16');
INSERT INTO @TABLE VALUES ('02/28/2017 13:45','02/28/2017 23:59','00:17');

UPDATE @TABLE
SET Duration = Duration + (Finish - Start);

返回：

Start                    Finish                 Duration
---------------------------------------------------------   
27/02/2017 08:00:00      28/02/2017 08:30:00    00:45:00
28/02/2017 09:00:00      28/02/2017 09:40:00    00:54:00
28/02/2017 10:02:00      28/02/2017 11:53:00    02:04:00
28/02/2017 11:56:00      28/02/2017 12:45:00    03:05:00
28/02/2017 13:45:00      28/02/2017 23:59:00    10:31:00

Answer 3

UPDATE [TABLE] SET [DURATION] = CONVERT（varchar（5），DATEADD（分钟，DATEDIFF（MINUTE，StartDate，EndDate），0），114）;

Answer 4

其他任何一个答案都没有尝试使你期望的持续时间相加。

DECLARE @TABLE TABLE (ID INT, StartTime DATETIME, EndTime DATETIME, Duration TIME);

INSERT INTO @TABLE VALUES (1, '02/28/2017 01:00','02/28/2017 06:30','');
INSERT INTO @TABLE VALUES (2, '02/28/2017 09:00','02/28/2017 23:40','');
INSERT INTO @TABLE VALUES (3, '03/01/2017 10:02','03/01/2017 21:53','');

UPDATE t
SET Duration=
    CONVERT(TIME,
        (
            SELECT DATEADD(ms, SUM(DATEDIFF(ms, '00:00:00.000', EndTime - StartTime)), '00:00:00.000')
            FROM @TABLE it
            WHERE it.EndTime<= t.EndTime
        )
    )
FROM @TABLE t;

SELECT * FROM @TABLE;

如何在SQL Server中添加time数据类型列

4 个答案: