Deterministic scalar function to get day of week for a date

Question

SQL Server, trying to get day of week via a deterministic UDF.

Im sure this must be possible, but cant figure it out.

UPDATE: SAMPLE CODE..

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Answer 1

Although this is old. Commenting simple answer to help future readers. @liver-larson was almost correct. Made a wrong assumption of epoch time 0 (Jan 1st, 1900) being a Sunday. It is actually a Monday.

Wrong code - https://stackoverflow.com/a/20481717/2012977

datediff(dd,0,[DateColumn]) % 7 + 1

Correction

datediff(dd,-1,[DateColumn]) % 7 + 1

I like this answer because there is no date conversion happening and guaranteed deterministic.

Unit test of code. Sadly SqlFiddle down at the moment of writing. Note hardcoding column value here. Expect column to be a date type.

SELECT DATEPART(WEEKDAY, '2020-06-14'), DATEDIFF(dd, -1, '2020-06-14') % 7 + 1
SELECT DATEPART(WEEKDAY, '2020-06-15'), DATEDIFF(dd, -1, '2020-06-15') % 7 + 1
SELECT DATEPART(WEEKDAY, '2020-06-16'), DATEDIFF(dd, -1, '2020-06-16') % 7 + 1
SELECT DATEPART(WEEKDAY, '2020-06-17'), DATEDIFF(dd, -1, '2020-06-17') % 7 + 1
SELECT DATEPART(WEEKDAY, '2020-06-18'), DATEDIFF(dd, -1, '2020-06-18') % 7 + 1
SELECT DATEPART(WEEKDAY, '2020-06-19'), DATEDIFF(dd, -1, '2020-06-19') % 7 + 1
SELECT DATEPART(WEEKDAY, '2020-06-20'), DATEDIFF(dd, -1, '2020-06-20') % 7 + 1