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sql - 获取每周的开始日期和结束日期

问题描述

这是我的代码

declare @startdate date='2021-01-01'
declare @enddate date='2021-12-31'
declare @week_end_date date

while (@startdate <= @enddate)
begin
    if (datepart(week, @startdate) = 1)
        select '01-01-2021'
    else
        set @startdate = DATEADD(day,1-datepart(dw,@startdate),@startdate)

    select @startdate week_start_date

    if (datepart(week, @startdate) = 53)
        select '2021-12-31' 
    else
        set @week_end_date = dateadd(day, 7 - datepart(dw, @startdate), @startdate)

    select @week_end_date week_end_date

    set @startdate = dateadd(week, 1, @startdate)   
end

我想达到以下结果

week_start date             week_end_date
-----------------------------------------
2021-01-01                   2021-01-02
2021-01-03                   2021-01-09
2021-01-10                   2021-01-16
.
.
.
.
.
.
2021-12-26                  2021-12-31
.
.

我正在尝试获取start_date每周end_date的一周，但我希望第一周的开始日期和上周的2020-01-01结束日期为 2020-12-31

在这两种情况下else都在运行。

标签： sqlsql-servertsql

解决方案

我猜你真的只需要涵盖一个完整的日历年。前两个 CTE 只是创建一个“数字表”。如有必要，您可以插入替换件。我也假设datefirst设置为 7。

with d as (
    select n from (values (0), (1), (2), (3), (4), (5), (6), (7)) t(n)
), w as (
    select d0.n + d1.n * 8 as offset from d as d0 cross join d as d1
), weeks as (
    select
        offset + 1 as weeknum,
        dateadd(week, offset, basedate) as weekstart,
        dateadd(day, 6, dateadd(week, offset, basedate)) as weekend
    from w cross apply (
        select dateadd(day, 1 - datepart(weekday, @startdate), @startdate)
    ) as v(basedate)
)
select
    weeknum,
    case when weekstart < @startdate then @startdate else weekstart end,
    case when weekend   > @enddate   then @enddate   else weekend end
from weeks
where weekstart <= @enddate;

https://dbfiddle.uk/?rdbms=sqlserver_2014&fiddle=5f900ecd26b92e4a112f191730958cae

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