DateDiff в SQL Server
SELECT DATEDIFF(dd, '2017/08/25', '2011/08/25') AS DateDiff
--result
-2192
Kalineo
SELECT DATEDIFF(dd, '2017/08/25', '2011/08/25') AS DateDiff
--result
-2192DATEDIFF(DAY, '1/1/2011', '3/1/2011')SELECT t1.OrderNo,DATEDIFF(day,t1.LoadedStartDate,t2.LoadedStartDate)
FROM UnnamedTableFromQuestion t1
INNER JOIN
UnnamedTableFromQuestion t2
on
t1.OrderNo = t2.OrderNo
WHERE t1.OpNo = 1 and
t2.OpNo = 4
////////////////////////////////////////////////////////////////
select cur.unique_id_field, cur.seq_no, cur.date_created ,
datediff(second, prv.date_created, cur.date_created) as diff_in_seconds
from yourtable as cur
join yourtable as prv
on cur.seq_no = prv.seq_no + 1;CREATE FUNCTION FullMonthsSeparation
(
@DateA DATETIME,
@DateB DATETIME
)
RETURNS INT
AS
BEGIN
DECLARE @Result INT
DECLARE @DateX DATETIME
DECLARE @DateY DATETIME
IF(@DateA < @DateB)
BEGIN
SET @DateX = @DateA
SET @DateY = @DateB
END
ELSE
BEGIN
SET @DateX = @DateB
SET @DateY = @DateA
END
SET @Result = (
SELECT
CASE
WHEN DATEPART(DAY, @DateX) > DATEPART(DAY, @DateY)
THEN DATEDIFF(MONTH, @DateX, @DateY) - 1
ELSE DATEDIFF(MONTH, @DateX, @DateY)
END
)
RETURN @Result
END
GO
SELECT dbo.FullMonthsSeparation('2009-04-16', '2009-05-15') as MonthSep -- =0
SELECT dbo.FullMonthsSeparation('2009-04-16', '2009-05-16') as MonthSep -- =1
SELECT dbo.FullMonthsSeparation('2009-04-16', '2009-06-16') as MonthSep -- =2