select cast(left('200210250125',4)+'-'+substring('200210250125',5,2)+'-'+substring('200210250125',7,2)+' '+left(right('200210250125',4),2)+':'+right('200210250125',2)+':00' as datetime)
select cast(left('200210250125',4)+'-'+substring('200210250125',5,2)+'-'+substring('200210250125',7,2)+' '+left(right('200210250125',4),2)+':'+right('200210250125',2)+':00' as datetime)
set @a='200210250125'
select convert(char(10),cast(substring(@a,1,8) as datetime),120)+' ' +substring(@a,9,2)+':'+substring(@a,11,2)select cast((convert(char(10),cast(substring(@a,1,8) as datetime),120)+' ' +substring(@a,9,2)+':'+substring(@a,11,2)) as datetime)
where datediff(m,time_field,getdate())<5
where
datediff(m, convert(datetime,left(tiem,8)+' ' +substring(tiem,9,2)+':'+right(tiem,2)),getdate())<5
请问如果这里我的time_field是从某字符窜类型的字段(如200212250125),这里需要转换为时间,应该怎么半??谢谢你
time_field字段如何转换为日期???
请帮我看一下,这样能够选出近5分钟内的记录吗,请问一下,如果一个是200312250125,当前系统时间是200312271225,这样得到的差值是一个很大的数字(代表相差的分钟数吗?)谢谢