sql-server-2008 - SQL Server - calculate elapsed time between two datetime stamps in HH:MM:SS format

Question

I have a SQL Server table that has a "Time" column. The table is a log table the houses status messages and timestamps for each message. The log table is inserted into via a batch file. There is an ID column that groups rows together. Each time the batch file runs it initializes the ID and writes records. What I need to do is get the elapsed time from the first record in an ID set to the last record of the same ID set. I started toying with select Max(Time) - Min(Time) from logTable where id = but couldn't figure out how to format it correctly. I need it in HH:MM:SS.

Answer 1

更新：

正确计算 SQL Server 中的时间跨度，即使超过 24 小时：

-- Setup test data
declare @minDate datetime = '2012-12-12 20:16:47.160'
declare @maxDate datetime = '2012-12-13 15:10:12.050'

-- Get timespan in hh:mi:ss
select cast(
        (cast(cast(@maxDate as float) - cast(@minDate as float) as int) * 24) /* hours over 24 */
        + datepart(hh, @maxDate - @minDate) /* hours */
        as varchar(10))
    + ':' + right('0' + cast(datepart(mi, @maxDate - @minDate) as varchar(2)), 2) /* minutes */
    + ':' + right('0' + cast(datepart(ss, @maxDate - @minDate) as varchar(2)), 2) /* seconds */

-- Returns 18:53:24

特别欢迎显示不准确的边缘案例！

Answer 2

SQL Server 不支持 SQL 标准间隔数据类型。最好的办法是以秒为单位计算差异，并使用函数来格式化结果。只要您的时间间隔小于 24 小时，本机函数 CONVERT() 可能看起来工作正常。但是 CONVERT() 不是一个好的解决方案。

create table test (
  id integer not null,
  ts datetime not null
  );

insert into test values (1, '2012-01-01 08:00');
insert into test values (1, '2012-01-01 09:00');
insert into test values (1, '2012-01-01 08:30');
insert into test values (2, '2012-01-01 08:30');
insert into test values (2, '2012-01-01 10:30');
insert into test values (2, '2012-01-01 09:00');
insert into test values (3, '2012-01-01 09:00');
insert into test values (3, '2012-01-02 12:00');

值的选择方式使得

• id = 1，经过的时间是1小时
• id = 2，经过的时间是 2 小时，并且
• id = 3，经过的时间是 3 小时。

此 SELECT 语句包括一个计算秒数的列，以及一个使用 CONVERT() 减法的列。

select t.id,
       min(ts) start_time,
       max(ts) end_time,
       datediff(second, min(ts),max(ts)) elapsed_sec,
       convert(varchar, max(ts) - min(ts), 108) do_not_use
from test t
group by t.id;

ID  START_TIME                 END_TIME                   ELAPSED_SEC  DO_NOT_USE
1   January, 01 2012 08:00:00  January, 01 2012 09:00:00  3600         01:00:00
2   January, 01 2012 08:30:00  January, 01 2012 10:30:00  7200         02:00:00
3   January, 01 2012 09:00:00  January, 02 2012 12:00:00  97200        03:00:00

请注意 3 号 27 小时时差的误导性“03:00:00”。

在 SQL Server 中格式化经过时间的函数

Answer 3

DECLARE @EndTime AS DATETIME, @StartTime AS DATETIME

SELECT @StartTime = '2013-03-08 08:00:00', @EndTime = '2013-03-08 08:30:00'

SELECT CAST(@EndTime - @StartTime AS TIME)

结果：00:30:00.0000000

按照您认为合适的方式格式化结果。

Answer 4

最好和最简单的方法：

Convert(varchar, {EndTime} - {StartTime}, 108)

就像杏里所说的那样。

Answer 5

使用 DATEDIFF 以毫秒、秒、分钟、小时、...为单位返回值

DATEDIFF（时间间隔，日期 1，日期 2）

interval REQUIRED - 要返回的时间/日期部分。可以是以下值之一：

year, yyyy, yy = Year
quarter, qq, q = Quarter
month, mm, m = month
dayofyear = Day of the year
day, dy, y = Day
week, ww, wk = Week
weekday, dw, w = Weekday
hour, hh = hour
minute, mi, n = Minute
second, ss, s = Second
millisecond, ms = Millisecond

date1, date2 REQUIRED - 计算两者之间差异的两个日期

Answer 6

看看这是否有帮助。我可以为经过的天数、小时、分钟、秒设置变量。您可以根据自己的喜好对其进行格式化或包含在用户定义的函数中。

注意：不要使用 DateDiff(hh,@Date1,@Date2)。这不可靠！它以不可预知的方式循环

给定两个日期...（示例日期：两天，三小时，10 分钟，30 秒差异）

declare @Date1 datetime = '2013-03-08 08:00:00'
declare @Date2 datetime = '2013-03-10 11:10:30'
declare @Days decimal
declare @Hours decimal
declare @Minutes decimal
declare @Seconds decimal

select @Days = DATEDIFF(ss,@Date1,@Date2)/60/60/24 --Days
declare @RemainderDate as datetime = @Date2 - @Days
select @Hours = datediff(ss, @Date1, @RemainderDate)/60/60 --Hours
set @RemainderDate = @RemainderDate - (@Hours/24.0)
select @Minutes = datediff(ss, @Date1, @RemainderDate)/60 --Minutes
set @RemainderDate = @RemainderDate - (@Minutes/24.0/60)
select @Seconds = DATEDIFF(SS, @Date1, @RemainderDate)    
select @Days as ElapsedDays, @Hours as ElapsedHours, @Minutes as ElapsedMinutes, @Seconds as ElapsedSeconds

Answer 7

select convert(varchar, Max(Time) - Min(Time) , 108) from logTable where id=...

Answer 8

希望这可以帮助您获得两个时间戳之间的确切时间

Create PROC TimeDurationbetween2times(@iTime as time,@oTime as time) 
As  
Begin  

DECLARE @Dh int, @Dm int, @Ds int ,@Im int, @Om int, @Is int,@Os int     

SET @Im=DATEPART(MI,@iTime)  
SET @Om=DATEPART(MI,@oTime)  
SET @Is=DATEPART(SS,@iTime)  
SET @Os=DATEPART(SS,@oTime)  

SET @Dh=DATEDIFF(hh,@iTime,@oTime)  
SET @Dm = DATEDIFF(mi,@iTime,@oTime)  
SET @Ds = DATEDIFF(ss,@iTime,@oTime)  

DECLARE @HH as int, @MI as int, @SS as int  

if(@Im>@Om)  
begin  
SET @Dh=@Dh-1  
end  
if(@Is>@Os)  
begin  
SET @Dm=@Dm-1  
end  

SET @HH = @Dh  
SET @MI = @Dm-(60*@HH)  
SET @SS = @Ds-(60*@Dm)  

DECLARE @hrsWkd as varchar(8)         

SET @hrsWkd = cast(@HH as char(2))+':'+cast(@MI as char(2))+':'+cast(@SS as char(2))          

select @hrsWkd as TimeDuration   

End