select convert(char(10),cast(substring('2012030607:37:54',0,9)as DATEtime),120)+'-'+RIGHT('2012030607:37:54',8)

------解决方案--------------------
SQL code


select convert(datetime,left('2012030607:37:54',8)+' '+right('2012030607:37:54',
LEN('2012030607:37:54')-8),120) as [date]
date
2012-03-06 07:37:54.000

------解决方案--------------------
习惯，在编码里面有些是以0开始的。

------解决方案--------------------
SQL code

select substring(date,1,4)+'-'+substring(date,5,2)+'-'+substring(date,7,2)+' '+substring(date,9,8) as date
from a
结果
date
----------------------
2012-03-06 07:37:54

(1 行受影响)

------解决方案--------------------
SQL code
create table tb(col varchar(16))
insert into tb values('2012030607:37:54')
insert into tb values('2012030607:38:38')
insert into tb values('2012030607:40:34')
insert into tb values('2012030607:41:10')
go

select left(col,4) + '-' + substring(col,5,2) + '-' + substring(col , 7,2) + ' ' + right(col,8) from tb

drop table tb

/*
                                    
----------------------------------- 
2012-03-06 07:37:54
2012-03-06 07:38:38
2012-03-06 07:40:34
2012-03-06 07:41:10

（所影响的行数为 4 行）

*/