这将给我一个基于日期的周数:
SELECT DATEPART(wk, '7/27/2016') AS [Week]
例如,返回 31。
现在,我需要做的是找到那个星期的第一天,并以短日期格式返回它。例如:
Given Week: 31
Return First Day of Week: July 24
或者
Given Week: 52
Return First Day of Week: Dec 25
我相信一周的默认第一天是星期日,这就是我需要的日期。
我在这里看到了几篇接近的帖子,但没有一篇让我一路走下去。
谢谢!
最佳答案
我帮助从内到外阅读这篇文章。我添加了带编号的评论以提供帮助。
declare @weekNum int;set @weeknum = 52;
select
-- 3. Add number of weeks
dateadd(wk, @weekNum,
--2. first day of week 0 for that year (may belong to previous year)
dateadd(ww, datediff(wk, 0,
--1. First date of the year (week 0)
dateadd(YEAR, datediff(year,0, getDate()),0)
),-1) -- -1 here because 1900-01-01 (date 0) was a Monday, and adding weeks to a Monday results in a Monday.
)
我们可以结合第二步和第三步,因为它们都会增加周数:
declare @weekNum int;set @weeknum = 52;
select
--2. first day of week 0 for that year (may belong to previous year) + number of weeks
dateadd(ww, @weekNum + datediff(wk, 0,
--1. First date of the year (week 0)
dateadd(YEAR, datediff(year,0, getDate()),0)
),-1) -- -1 here because 1900-01-01 (day 0) was a Monday. Adding weeks to a Monday results in a Monday
另外,我认为你的第 31 周的例子比现在晚了一周。您可以像这样查看当年的完整集合:
with weeks as
(
select top 52 row_number() over (order by object_id) as wk from sys.objects
)
select wk,
--2. first day of week 0 for that year (may belong to previous year) + number of weeks
dateadd(ww, wk + datediff(wk, 0,
--1. First date of the year (week 0)
dateadd(YEAR, datediff(year,0, getDate()),0)
),-1) -- -1 here because 1900-01-01 (day 0) was a Monday. Adding weeks to a Monday results in a Monday
from weeks
关于sql - 给定周数,在 T-SQL 中返回一周的第一天,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/38615906/