How to calculate no. of days between two datetime column in SQL Server - sql

Question

There are two tables, T1 includes start date column and T2 includes end date column. Both the columns have both date and time.
T1 has all the rows unique while T2 can have multiple rows for same id and in all the row the end date column might be different (each row with different date and time).
I want to calculate the difference (no. of days) between End date and the start date while keeping in mind that we have to only pick the last date which is lying in the End date column.

Answer 1

;WITH MaxEndDate AS
(
SELECT
T2.PrimaryKey,
MaxEndDate = MAX(EndDate)
FROM
T2
GROUP BY
T2.PrimaryKey
)
SELECT
T1.PrimaryKey,
T1.StartDate,
M.MaxEndDate,
AmountDays = DATEDIFF(DAY, T1.StartDate, M.MaxEndDate)
FROM
T1
INNER JOIN MaxEndDate AS M ON T1.PrimaryKey = M.PrimaryKey

Answer 2

I would recommend creating a temporary table using T2 something like
Select Distinct ID
,MAX(ENDDATE) As [ENDDATE]
INTO #TMPTABLE1
then you include this into T1
Select A.ID
,Startdate
,B.ENDDATE
,Datediff(day,A.STARTDATE,B.ENDDATE) as DAYS
From T1 as A inner join
#TEMPTABLE1 as B on A.ID = B.ID

Answer 3

--------get no of days among multiple dates----------
DECLARE #i INT , #numrows INT , #days INT , #Fdate DATETIME , #Tdate DATETIME;
SET
#days = 0;
DECLARE #employee_table TABLE ( idx SMALLINT PRIMARY KEY IDENTITY(1,
1) ,
EmpId INT ,
SinceDate DATETIME ,
TillDate DATETIME );
-- populate employee table INSERT #employee_table
SELECT
EmpId ,
SinceDate ,
TillDate
FROM
T_EmpPosting_PIS
WHERE
EmpId = 18
AND OfficeTypeID = 1
ORDER BY
TransDate;
--SELECT
*
FROM
#employee_table -- enumerate the table
SET
#i = 1;
SET
#numrows = ( SELECT
COUNT(*)
FROM
#employee_table );
IF #numrows > 0 WHILE ( #i <= ( SELECT
MAX(idx)
FROM
#employee_table ) ) BEGIN
SET
#Fdate = ( SELECT
SinceDate
FROM
#employee_table
WHERE
idx = #i );
IF ( #i = #numrows ) BEGIN
SET
#Tdate = GETDATE();
END;
ELSE BEGIN
SET
#Tdate = ( SELECT
TillDate
FROM
#employee_table
WHERE
idx = #i );
END;
SET
#days = ( SELECT
DATEDIFF(DAY,
#Fdate,
#Tdate) ) #days;
SET
#i = #i 1;
END;
SELECT
#days;