尽管SQLMenace的答案已经被接受，但是您应该注意一个重要的SET选项

设置DATEFIRST

如果一周的第一天发生了如下所示的变化，那么DATENAME将返回正确的日期名称，但不是相同的DATEPART值。

declare @DefaultDateFirst int
set @DefaultDateFirst = @@datefirst
--; 7 First day of week is "Sunday" by default
select  [@DefaultDateFirst] = @DefaultDateFirst 
    
set datefirst @DefaultDateFirst
select datename(dw,getdate()) -- Saturday
select datepart(dw,getdate()) -- 7

--; Set the first day of week to * TUESDAY * 
--; (some people start their week on Tuesdays...)
set datefirst 2
select datename(dw,getdate()) -- Saturday
--; Returns 5 because Saturday is the 5th day since Tuesday.
--; Tue 1, Wed 2, Th 3, Fri 4, Sat 5
select datepart(dw,getdate()) -- 5 <-- It's not 7!
set datefirst @DefaultDateFirst

尽管SQLMenace的答案已经被接受，但是您应该注意一个重要的SET选项

设置DATEFIRST

如果一周的第一天发生了如下所示的变化，那么DATENAME将返回正确的日期名称，但不是相同的DATEPART值。

declare @DefaultDateFirst int
set @DefaultDateFirst = @@datefirst
--; 7 First day of week is "Sunday" by default
select  [@DefaultDateFirst] = @DefaultDateFirst 
    
set datefirst @DefaultDateFirst
select datename(dw,getdate()) -- Saturday
select datepart(dw,getdate()) -- 7

--; Set the first day of week to * TUESDAY * 
--; (some people start their week on Tuesdays...)
set datefirst 2
select datename(dw,getdate()) -- Saturday
--; Returns 5 because Saturday is the 5th day since Tuesday.
--; Tue 1, Wed 2, Th 3, Fri 4, Sat 5
select datepart(dw,getdate()) -- 5 <-- It's not 7!
set datefirst @DefaultDateFirst

欧洲:

declare @d datetime;
set @d=getdate();
set @dow=((datepart(dw,@d) + @@DATEFIRST-2) % 7+1);

您可能会发现这个版本用于在一行中返回缩短的工作日名称。

-- Test DATA
select @@datefirst
create table #test (datum datetime)
insert #test values ('2013-01-01')
insert #test values ('2013-01-02')
insert #test values ('2013-01-03')
insert #test values ('2013-01-04')
insert #test values ('2013-01-05')
insert #test values ('2013-01-06')
insert #test values ('2013-01-07')
insert #test values ('2013-01-08')
-- Test DATA

select  Substring('Sun,Mon,Tue,Wed,Thu,Fri,Sat,Sun,Mon,Tue,Wed,Thu,Fri,Sat',
        (DATEPART(WEEKDAY,datum)+@@datefirst-1)*4+1,3),Datum
        from #test 

这是一个工作副本的我的代码检查它，如何从日期检索日期名称在SQL

CREATE Procedure [dbo].[proc_GetProjectDeploymentTimeSheetData] 
@FromDate date,
@ToDate date

As 
Begin
select p.ProjectName + ' ( ' + st.Time +' '+'-'+' '+et.Time +' )' as ProjectDeatils,
datename(dw,pts.StartDate) as 'Day'
from 
ProjectTimeSheet pts 
join Projects p on pts.ProjectID=p.ID 
join Timing st on pts.StartTimingId=st.Id
join Timing et on pts.EndTimingId=et.Id
where pts.StartDate >= @FromDate
and pts.StartDate <= @ToDate
END

如果您不想依赖于@@DATEFIRST或使用DATEPART(weekday, DateColumn)，只需自己计算星期几即可。

对于以周一为基准的周(欧洲)，最简单的方法是:

SELECT DATEDIFF(day, '17530101', DateColumn) % 7 + 1 AS MondayBasedDay

对于星期日为基础的周(美国)使用:

SELECT DATEDIFF(day, '17530107', DateColumn) % 7 + 1 AS SundayBasedDay

这返回了自1753年1月1日以来的工作日数字(1到7)。