请参考如下SQL，具体逻辑是：差值的绝对值小到大排序，取第一行即可。
如果需要其他列的值，把取值放where中…… 这样就算有重复值（差值的绝对值相同）也能揪出来~
select T from table(select T, abs($T_STR-T) as diff from A) order by diff fetch first 1 rows only
例子：
db2 => select * from a
T
--------------------------
2014-04-02-01.00.00.000000
2014-05-02-01.00.00.000000
2014-06-02-01.00.00.000000
3 record(s) selected.
db2 => values timestamp('2014-05-03-01.00.00.00000')
1
--------------------------
2014-05-03-01.00.00.000000
1 record(s) selected.
db2 => select T from table(select T, abs('2014-05-03-01.00.00.00000'-T) as diff from A) order by diff fetch first 1 rows only
T
--------------------------
2014-05-02-01.00.00.000000
1 record(s) selected.
db2 =>
