how can i know the total number of holidays by employee - sql-server

Question

I have this tables Holiday(Id,FK(EmployeeId),StartDate,EndDate) and table Employee(Id,FullName,etc...)
I want to know the number of days that each employee have
I was trying something like this :
SELECT Employee.Id, SUM(DATEDIFF(day,Holiday.StartDate,Holiday.EndDate) + 1)
FROM Employee
LEFT JOIN Holiday ON Holiday.EmployeeId=Employee.Id
GROUP BY Employee.id
i know this doesn't work because, to sum thati would need to group by Holiday.Id since i will have many rows in the Holiday table for the same EmployeeId
how can i accomplish this?
thanks for the help

Answer 1

Or, using a workingday calculation I found here: https://www.sqlshack.com/how-to-calculate-work-days-and-hours-in-sql-server you could do the following:
CREATE FUNCTION workingdays ( #DateFrom Date, #DateTo Date) RETURNS INT AS
BEGIN
DECLARE #TotDays INT= DATEDIFF(DAY, #DateFrom, #DateTo) + 1;
DECLARE #TotWeeks INT= DATEDIFF(WEEK, #DateFrom, #DateTo) * 2;
DECLARE #IsSunday INT= CASE
WHEN DATENAME(WEEKDAY, #DateFrom) = 'Sunday'
THEN 1
ELSE 0
END;
DECLARE #IsSaturday INT= CASE
WHEN DATENAME(WEEKDAY, #DateTo) = 'Saturday'
THEN 1
ELSE 0
END;
DECLARE #TotWorkingDays INT= #TotDays - #TotWeeks - #IsSunday + #IsSaturday;
RETURN #TotWorkingDays;
END
GO
create table Employee (Id int identity, name varchar(64), Primary Key (Id));
create table Holiday (EmployeeId int, StartDate date, EndDate date);
insert into Employee VALUES ('Harry Potter'),('Hermiony Granger'),('Ron Weasly'),('Ginny Weasley');
insert into Holiday VALUES (1,'2020-02-12','2020-02-18'),(1,'2020-04-02','2020-04-07'),(1,'2020-08-21','2020-09-05'),
(2,'2020-01-04','2020-01-13'),(2,'2020-03-17','2020-03-23'),(2,'2020-05-29','2020-06-7');
SELECT Employee.Id, SUM(dbo.workingdays(Holiday.StartDate,Holiday.EndDate))
FROM Employee
LEFT JOIN Holiday ON Holiday.EmployeeId=Employee.Id
GROUP BY Employee.id
This will still only be a crude estimation as is does not account for public holidays.
DEMO: https://rextester.com/QKGUT41272

Answer 2

Hi i think you can build query using CTE like this :
Resource : CTE Microsoft : https://learn.microsoft.com/fr-fr/sql/t-sql/queries/with-common-table-expression-transact-sql?view=sql-server-ver15
Or using select in clause form
EDIT : Or it work with your current query to
create Table #EMPLOYEE
(
EMP_ID INT,
EMP_NAME varchar(128)
)
create Table #HOLIDAYS
(
HL_ID INT,
HL_EMP varchar(128),
StartDate DATE,
EndDate DATE
)
Insert into #EMPLOYEE
(
EMP_ID,
EMP_NAME
)
SELECT 1, 'Toto'
UNION ALL
SELECT 2,'Dupont'
UNION ALL
SELECT 3,'Titi'
UNION ALL
SELECT 4,'Tata'
Insert into #HOLIDAYS
(
HL_ID,
HL_EMP,
StartDate,
EndDate
)
SELECT '1', '1',GETDATE(),DATEADD(day,4,GETDATE())
UNION ALL
SELECT '2','1',DATEADD(day,-7,GETDATE()), DATEADD(day,-5,GETDATE())
UNION ALL
SELECT '3','2',DATEADD(day,4,GETDATE()),DATEADD(day,15,GETDATE())
-- USING CTE EXEMPLE
;WITH MyCteAPP AS (
SELECT EMP_ID, EMP_NAME, HL_ID, ISNULL(StartDate,GETDATE()) AS 'StartDate', ISNULL(EndDate,GETDATE()) AS 'EndDate'
FROM #EMPLOYEE
LEFT JOIN #HOLIDAYS ON EMP_ID = HL_EMP
)
SELECT EMP_ID, EMP_NAME, SUM(DATEDIFF(day,StartDate,EndDate)) AS 'NbDay'
FROM MyCteAPP
GROUP BY EMP_ID,EMP_NAME
ORDER BY EMP_ID
-- USING SELECT IN FROM EXEMPLE
SELECT EMP_ID, EMP_NAME, SUM(DATEDIFF(day,StartDate,EndDate)) AS 'NbDay'
FROM (SELECT EMP_ID, EMP_NAME, HL_ID, ISNULL(StartDate,GETDATE()) AS 'StartDate', ISNULL(EndDate,GETDATE()) AS 'EndDate'
FROM #EMPLOYEE
LEFT JOIN #HOLIDAYS ON EMP_ID = HL_EMP
) AS SUBQUERY
GROUP BY EMP_ID,EMP_NAME
ORDER BY EMP_ID
--USING CURRENT QUERY
SELECT EMP_ID, EMP_NAME, SUM(DATEDIFF(day,ISNULL(StartDate,GETDATE()),ISNULL(EndDate,GETDATE()))) AS 'NbDay'
FROM #EMPLOYEE
LEFT JOIN #HOLIDAYS ON EMP_ID = HL_EMP
GROUP BY EMP_ID,EMP_NAME
ORDER BY EMP_ID
DROP TABLE #EMPLOYEE
DROP TABLE #HOLIDAYS
RESULT :

Answer 3

I have tried with sub query. Please use below query. It will be helpful.
SELECT Id, SUM(leave) AS leave
FROM
(
SELECT Employee.Id, DATEDIFF(dd,Holiday.StartDate,Holiday.EndDate) as leave
FROM Employee
LEFT JOIN Holiday ON Holiday.EmployeeId=Employee.Id
)a
GROUP BY ID