select top 1 * from tb
order by abs(datediff(d,time,@time)) asc

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

declare @T table (time datetime,empname varchar(4),proId int)
insert into @T
select '2011-12-01','张三',1 union all
select '2011-12-01','张三',2 union all
select '2012-01-01','张三',3 union all
select '2012-01-01','张三',4 union all
select '2012-01-01','张三',5 union all
select '2011-12-01','李四',1 union all
select '2011-12-01','李四',2

declare @date varchar(10) set @date='2012-02-01'

select * from @T t where abs(datediff(d,time,@date))
=(select min(abs(datediff(d,time,@date))) from @T where empname=t.empname)
/*
time                    empname proId
----------------------- ------- -----------
2011-12-01 00:00:00.000 李四      1
2011-12-01 00:00:00.000 李四      2
2012-01-01 00:00:00.000 张三      3
2012-01-01 00:00:00.000 张三      4
2012-01-01 00:00:00.000 张三      5
*/

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

DECLARE @EndDate DATETIME
SET @EndDate = '2012-02-01'
declare @T table (time datetime,empname varchar(4),proId int)
insert into @T
select '2011-12-01','张三',1 union all
select '2011-12-01','张三',2 union all
select '2012-01-01','张三',3 union all
select '2012-01-01','张三',4 union all
select '2012-01-01','张三',5 union all
select '2011-12-01','李四',1 union all
select '2011-12-01','李四',2

SELECT * FROM @T A
WHERE time = (SELECT TOP 1 time FROM @T WHERE empname = A.empname AND time <=@EndDate ORDER BY time DESC)