DECLARE @a varchar(14) ='20120907112032'
SELECT CONVERT(datetime,SUBSTRING(@a,1,4)+'-'+SUBSTRING(@a,5,2)+'-'+SUBSTRING(@a,7,2)+' '+SUBSTRING(@a,9,2)+':'+SUBSTRING(@a,11,2)+':'+SUBSTRING(@a,13,2))

------解决方案--------------------

SQL code
declare @a char(14)
declare @aa varchar(25)

set @a='20120907112032'
set @aa=left(@a,4)+'-'+right(left(@a,6),2)+'-'+right(left(@a,8),2)+' '+right(left(@a,10),2)+':'+right(left(@a,12),2)+':'+right(@a,2)+'.000'

select cast(@aa as datetime) as '时间'

/*
-->测试结果
时间                          
--------------------------- 
2012-09-07 11:20:32.000

*/

------解决方案--------------------

SQL code
DECLARE @a varchar(14) ='20120907112032'
SELECT CONVERT(datetime,SUBSTRING(@a,1,4)+'-'+SUBSTRING(@a,5,2)+'-'
+SUBSTRING(@a,7,2)+' '+SUBSTRING(@a,9,2)+':'+SUBSTRING(@a,11,2)+
':'+SUBSTRING(@a,13,2))
------------------------------
/*
无列名
2012-09-07 11:20:32.000
*/

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

declare @a char(14)
set @a='20120907112032'
select convert(varchar(10),convert(datetime,left(@a,8)),120)
    +' '+SUBSTRING(@a,9,2)+':'+SUBSTRING(@a,11,2)+':'+SUBSTRING(@a,13,2)
/*
2012-09-07 11:20:32
*/

------解决方案--------------------
探讨

SQL code

declare @a char(14)
set @a='20120907112032'
select convert(varchar(10),convert(datetime,left(@a,8)),120)
    +' '+SUBSTRING(@a,9,2)+':'+SUBSTRING(@a,11,2)+':'+SUBSTRING(@a,13,2)
/*
20……


------解决方案--------------------
尝试了函数直接转换，没成功。

------解决方案--------------------
探讨


尝试了函数直接转换，没成功。