DB2中，如何查询出与某个时间最接近的记录

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

--------------------------

2014-04-02-010000000000

2014-05-02-010000000000

2014-06-02-010000000000

3 record(s) selected

db2 => values timestamp('2014-05-03-01000000000')

1

--------------------------

2014-05-03-010000000000

1 record(s) selected

db2 => select T from table(select T, abs('2014-05-03-01000000000'-T) as diff from A) order by diff fetch first 1 rows only

T

--------------------------

2014-05-02-010000000000

1 record(s) selected

db2 =>

SELECT DATE(LEFT('201202', 4) || '-'||RIGHT('201202', 2)||'-1') - 1 MONTH,

DATE(LEFT('201202', 4) || '-'||RIGHT('201202', 2)||'-1') + 1 MONTH

from sysibm/sysdummy1

DATE(LEFT('201202', 4) || '-'||RIGHT('201202', 2)||'-1') - 1 MONTH这样拿到的是日期类型，需要cast as char,然后截取就可以了

db2处理日期很垃圾的

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