确定性标量函数，用于获取日期的一年中的第几周 [英] Deterministic scalar function to get week of year for a date

问题描述

这里是如何获取星期几的好方法，确定性标量函数获取日期的星期几.

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.

现在，有人可以帮我创建一个确定性标量函数来获取一年中的一周作为日期吗?谢谢.

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

推荐答案

这是确定性的，我可以将其用作计算列.

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

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

测试代码:

;
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)

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