如何比较两个日期以查找SQL Server 2005中的时差，日期操作

我有两列：

job_start                         job_end
2011-11-02 12:20:37.247           2011-11-02 13:35:14.613

如何使用T-SQL查找作业开始到结束之间经过的原始时间？

我尝试了这个：

select    (job_end - job_start) from tableA

但最终结果是：

1900-01-01 01:14:37.367

看一下DateDiff()功能。

-- Syntax
-- DATEDIFF ( datepart , startdate , enddate )

-- Example usage
SELECT DATEDIFF(DAY, GETDATE(), GETDATE() + 1) AS DayDiff
SELECT DATEDIFF(MINUTE, GETDATE(), GETDATE() + 1) AS MinuteDiff
SELECT DATEDIFF(SECOND, GETDATE(), GETDATE() + 1) AS SecondDiff
SELECT DATEDIFF(WEEK, GETDATE(), GETDATE() + 1) AS WeekDiff
SELECT DATEDIFF(HOUR, GETDATE(), GETDATE() + 1) AS HourDiff
...

您可以在操作中看到它/在这里玩

您可以使用DATEDIFF函数获取分钟，秒，天等之间的差异。

SELECT DATEDIFF(MINUTE,job_start,job_end)

MINUTE显然以分钟为单位返回时差，您也可以使用DAY，HOUR，SECOND，YEAR（有关完整列表，请参见在线图书链接）。

如果您想花哨的话，可以用不同的方式显示，例如75分钟可以这样显示：01：15：00：0

这是针对SQL Server 2005和2008执行此操作的代码

-- SQL Server 2005
SELECT CONVERT(VARCHAR(10),DATEADD(MINUTE,DATEDIFF(MINUTE,job_start,job_end),'2011-01-01 00:00:00.000'),114)

-- SQL Server 2008
SELECT CAST(DATEADD(MINUTE,DATEDIFF(MINUTE,job_start,job_end),'2011-01-01 00:00:00.000') AS TIME)

将结果强制转换为TIME，结果将以时间格式显示间隔的持续时间。

select CAST(job_end - job_start) AS TIME(0)) from tableA

我认为您需要job_start和job_end之间的时间间隔。

尝试这个...

select SUBSTRING(CONVERT(VARCHAR(20),(job_end - job_start),120),12,8) from tableA

我结束了这个。

01:14:37

声明开始和结束日期 DECLARE @SDATE AS DATETIME

TART_DATE  AS DATETIME
DECLARE @END_-- Set Start and End date
SET @START_DATE = GETDATE()
SET @END_DATE    = DATEADD(SECOND, 3910, GETDATE())

-以HH：MI：SS：MMM（24H）格式获取结果 SELECT CONVERT(VARCHAR(12), DATEADD(MS, DATEDIFF(MS, @START_DATE, @END_DATE), 0), 114) AS TimeDiff

在Sql Server中尝试

SELECT 
      start_date as firstdate,end_date as seconddate
       ,cast(datediff(MI,start_date,end_date)as decimal(10,3)) as minutediff
      ,cast(cast(cast(datediff(MI,start_date,end_date)as decimal(10,3)) / (24*60) as int ) as varchar(10)) + ' ' + 'Days' + ' ' 
      + cast(cast((cast(datediff(MI,start_date,end_date)as decimal(10,3)) / (24*60) - 
        floor(cast(datediff(MI,start_date,end_date)as decimal(10,3)) / (24*60)) ) * 24 as int) as varchar(10)) + ':' 

     + cast( cast(((cast(datediff(MI,start_date,end_date)as decimal(10,3)) / (24*60) 
      - floor(cast(datediff(MI,start_date,end_date)as decimal(10,3)) / (24*60)))*24
        -
        cast(floor((cast(datediff(MI,start_date,end_date)as decimal(10,3)) / (24*60) 
      - floor(cast(datediff(MI,start_date,end_date)as decimal(10,3)) / (24*60)))*24) as decimal)) * 60 as int) as varchar(10))

    FROM [AdventureWorks2012].dbo.learndate

如果您的数据库StartTime = 07:00:00和endtime = 14:00:00，并且都是时间类型。您的查询以获取时差为：

SELECT TIMEDIFF(Time(endtime ), Time(StartTime )) from tbl_name

如果数据库startDate =2014-07-20 07:00:00和endtime = 2014-07-20 23:00:00，则也可以使用此查询。

下面的代码以hh：mm格式给出。

选择RIGHT（LEFT（job_end- job_start，17），5）

看看DATEDIFF，这应该是您想要的。它需要您比较的两个日期，以及您想要的差异的日期单位（天，月，秒...）

我使用以下逻辑，对我来说就像奇迹一样有效：

CONVERT(TIME, DATEADD(MINUTE, DATEDIFF(MINUTE, AP.Time_IN, AP.Time_OUT), 0)) 

如果您尝试以一定的准确性获得工作时间，请尝试使用此方法（已在SQL Server 2016中测试）

SELECT DATEDIFF(MINUTE,job_start, job_end)/60.00;

各种DATEDIFF功能是：

SELECT DATEDIFF(year,        '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(quarter,     '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(month,       '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(dayofyear,   '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(day,         '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(week,        '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(hour,        '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(minute,      '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(second,      '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(millisecond, '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');

参考：https : //docs.microsoft.com/zh-cn/sql/t-sql/functions/datediff-transact-sql? view = sql-server- 2017