从oracle sql列表中获取最接近的最高值
我在数据库中有一列,格式如下:yymmddh24miss从oracle sql列表中获取最接近的最高值,sql,oracle,logic,Sql,Oracle,Logic,我在数据库中有一列,格式如下:yymmddh24miss Sample Data: 140203101241 140202101141 140102101240 143001101244 142801101245 142701131347 142601121542 142101131744 ... 我需要从列表中获取最近的高值。如果我通过了142701131333,那么它应该从上面的列表返回142701131347 感谢您的帮助 SELECT data FROM ( SELECT dat
Sample Data:
140203101241
140202101141
140102101240
143001101244
142801101245
142701131347
142601121542
142101131744
...
我需要从列表中获取最近的高值。如果我通过了142701131333,那么它应该从上面的列表返回142701131347
感谢您的帮助
SELECT data
FROM
(
SELECT data
FROM tbl
WHERE data > '142701131333'
ORDER BY data
) a
WHERE rownum = 1
SQL> with t (x) as (
2 select 140203101241 from dual union all
3 select 140202101141 from dual union all
4 select 140102101240 from dual union all
5 select 143001101244 from dual union all
6 select 142801101245 from dual union all
7 select 142701131347 from dual union all
8 select 142601121542 from dual union all
9 select 142101131744 from dual
10 )
11 select min(x) minx from t where x > 142701131333
12 /
MINX
-----------------
142701131347
SELECT MIN(sample_data)
FROM tableName
WHERE sample_data > 142701131333