SQL只取日期的年月日部分

方法一:
select CONVERT(varchar, getdate(), 120 )
2004-09-12 11:06:08
select replace(replace(replace(CONVERT(varchar, getdate(), 120 ),\'-\',\'\'),\' \',\'\'),\':\',\'\')
20040912110608 h
select CONVERT(varchar(12) , getdate(), 111 )
2004/09/12
select CONVERT(varchar(12) , getdate(), 112 )
20040912
select CONVERT(varchar(12) , getdate(), 102 )
2004.09.12
select CONVERT(varchar(12) , getdate(), 101 )
09/12/2004
select CONVERT(varchar(12) , getdate(), 103 )
12/09/2004
select CONVERT(varchar(12) , getdate(), 104 )
12.09.2004
select CONVERT(varchar(12) , getdate(), 105 )
12-09-2004
select CONVERT(varchar(12) , getdate(), 106 )
12 09 2004
select CONVERT(varchar(12) , getdate(), 107 )
09 12, 2004
select CONVERT(varchar(12) , getdate(), 108 )
11:06:08
select CONVERT(varchar(12) , getdate(), 109 )
09 12 2004 1
select CONVERT(varchar(12) , getdate(), 110 )
09-12-2004
select CONVERT(varchar(12) , getdate(), 113 )
12 09 2004 1
select CONVERT(varchar(12) , getdate(), 114 )
11:06:08.177
方法二:
declare @tt varchar(4),@pp varchar(8) ,@ss varchar(4),@aa varchar(16)
set @tt=rtrim(cast(month(getdate()) as varchar(10)))
set @ss=rtrim(cast(day(getdate()) as varchar(10)))
select @pp=rtrim(cast(year(getdate()) as varchar(10)))
select @aa=@pp+@tt+@ss
select @aa
只显示年月,不显示日:select datepart(year,getdate())*100+datepart(month,getdate())
                   200811
 select year(OrderDate) as yy ,sum(FreightTotMon)as price from [Order] group by  year(OrderDate)
select year(OrderDate)as yy ,month(OrderDate) as mon, sum(FreightTotMon) as price from [Order] group by month(OrderDate),year(OrderDate)
select OrderDate as date, sum(FreightTotMon) as price from [Order] group by OrderDate
--select OrderDate as date, FreightTotMon as price from [Order]
--select convert(varchar(12),OrderDate,111) as date, sum(FreightTotMon)