SQL statement for weekday

Question

WEEKDAY problem...what i wrong?

Line 2 and line 3 error Weekday

SELECT a.id, a.redni_broj , a.radnik, isnull(convert(varchar(20), a.vrijemeodjave, 113), '') as vrijemeodjave, convert(varchar(20), b.vrijemeprijave, 113) as vrijemeprijave,
CASE WHEN WEEKDAY(b.vrijemeprijave) in (5,6) then 0 else (CASE WHEN (DATEDIFF(SECOND, b.vrijemeprijave, a.vrijemeodjave) / 3600) > 8 THEN 8 ELSE(DATEDIFF(SECOND, b.vrijemeprijave, a.vrijemeodjave) / 3600) END) end AS radni_sati,
CASE WHEN WEEKDAY(a.vrijemeodjave) in (5,6) then 0 else (CASE WHEN(DATEDIFF(SECOND, b.vrijemeprijave, a.vrijemeodjave) / 3600) > 8 THEN(DATEDIFF(SECOND, b.vrijemeprijave, a.vrijemeodjave) / 3600) - 8 ELSE NULL END) end AS prekovremeni
FROM(SELECT id, redni_broj, radnik, vrijemeodjave, ROW_Number() OVER(Partition By id ORDER BY vrijemeprijave) as RowNum
FROM dbo.prijava_radnika) a INNER JOIN(SELECT id, redni_broj, radnik, vrijemeprijave, (ROW_Number() OVER(Partition By id ORDER BY vrijemeprijave) - 1) as RowNumMinusOne FROM dbo.prijava_radnika) b ON a.id = b.id where 1 = 1

Manoj Bhoir · Answer

Hi, It is simple 'WEEKDAY' is not a recognized built-in function name. There is no any inbuild function in SQL Server with name WEEKDAY. If it is user-defined function then you have to check it for any execution error as we don't have direct access to your PC or you SQL Server instance it is difficult for us to identify what was there in you environment without understanding table structure you have used in your query. Check for WEEKDAY function as it is not default SQL function. Hope this will help you. :)

Goran Bibic · Answer

I am solve problem...thank you

SQL statement for weekday

Answers (2)