Deterministic scalar function to get week of year for a date

问题

Here is a great way of how to get day of week for a date, Deterministic scalar function to get day of week for a date.

Now, could anyone help me to create a deterministic scalar function to get week of year for a date please? Thanks.

回答1:

This works deterministically, I can use it as a computed column.

datediff(week, dateadd(year, datediff(year, 0, @DateValue), 0), @DateValue) + 1 

Test code:

;
with 
Dates(DateValue) as 
(
    select cast('2000-01-01' as date)
    union all 
    select dateadd(day, 1, DateValue) from Dates where DateValue < '2050-01-01'
)
select 
    year(DateValue) * 10000 + month(DateValue) * 100 + day(DateValue) as DateKey, DateValue,        
    datediff(day, dateadd(week, datediff(week, 0, DateValue), 0), DateValue) + 2 as DayOfWeek,
    datediff(week, dateadd(month, datediff(month, 0, DateValue), 0), DateValue) + 1 as WeekOfMonth,
    datediff(week, dateadd(year, datediff(year, 0, DateValue), 0), DateValue) + 1 as WeekOfYear
    from Dates option (maxrecursion 0)

来源：https://stackoverflow.com/questions/11432719/deterministic-scalar-function-to-get-week-of-year-for-a-date