也许按（公司和日期）按（源）排序行，以便01（如果存在）位于02或任何其他值之前，然后提取排名最高的行？
样本数据：

SQL> with test (company, report_date, source) as
  2    (select '01', 20071231, '01' from dual union all
  3     select '01', 20081231, '01' from dual union all
  4     select '01', 20081231, '02' from dual union all
  5     select '02', 20071231, '02' from dual union all
  6     select '02', 20081231, '02' from dual union all
  7     select '03', 20071231, '01' from dual union all
  8     select '03', 20071231, '02' from dual
  9    ),

查询从这里开始：

10  temp as
 11    (select company, report_date, source,
 12       row_number() over (partition by company, report_date order by source) rn
 13     from test
 14    )
 15  select company, report_date, source
 16  from temp
 17  where rn = 1;

COMPANY    REPORT_DATE SOURCE
---------- ----------- ----------
01            20071231 01
01            20081231 01
02            20071231 02
02            20081231 02
03            20071231 01

SQL>

如果我正确理解了你的问题，那么你需要所有具有source的最小值的行来对应company。

select
    company,
    "report date",
    source
from
    mytable t1
where
    source = (
        select min(source)
        from mytable t2
        where t1.company = t2.company
    )

如果你只想要那些，其中source是最小的，但是01或02你可以添加条件：

select
    company,
    report_date,
    source
from
    mytable t1
where
    source = (
        select min(source)
        from mytable t2
        where t1.company = t2.company
    )
and source in ('01', '02')