我有一个包含“时间”列的 SQL Server 表。该表是一个日志表,其中包含状态消息和每条消息的时间戳。日志表通过批处理文件插入。有一个 ID 列将行分组在一起。每次运行批处理文件时,它都会初始化 ID 并写入记录。我需要做的是获取从 ID 集中的第一条记录到同一 ID 集的最后一条记录所经过的时间。我开始玩弄 select Max(Time) - Min(Time) from logTable where id = 但无法弄清楚如何正确格式化它。我需要 HH:MM:SS。
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。只要您的间隔小于 24 小时,本机函数 CONVERT() 可能看起来就可以正常工作。但 CONVERT() 并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');
选择值的方式使得
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 ID 号 3 上的 27 小时差异的误导性“03:00:00”。
更新:
在SQL Server中正确计算一个时间跨度,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!
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
Convert(varchar, {EndTime} - {StartTime}, 108)
就像安里所说的那样。
DATEDIFF(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 - 要计算之间差异的两个日期
select convert(varchar, Max(Time) - Min(Time) , 108) from logTable where id=...
注意:不要使用 DateDiff(hh,@Date1,@Date2)。这不可靠!它以不可预测的方式循环
给定两个日期... (日期示例: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
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
Declare @StartDate DateTime,@EndDate Datetime,
@seconds int,@secondsb int,
@minutes int, @minutesB int,
@Hours int,@hoursB int,
@Days int, @Daysb int,
@Years int;
set @StartDate='2021-01-10 10:12:13.987'
set @EndDate=getdate() --You can enter end date in above similar manners or use getdate() to compare current date and time
set @Seconds = DATEDIFF(SECOND, @StartDate, @EndDate)
set @secondsb=@Seconds%60
set @Minutes=(@Seconds-@SecondsB)/60
set @MinutesB=@Minutes%60
set @Hours=(@Minutes-@MinutesB)/60
set @HoursB=@Hours%24
set @days=(@Hours-@HoursB)/24
set @DaysB=@Days%365
set @Years=(@Days-@DaysB)/365
Select @Years Years, @Daysb Days, @hoursB Hours,@minutesB Minutes, @secondsb Seconds