可以使用解析COUNT函数（而不是聚合函数）：

select case
       when COUNT(a_code) OVER () > 1
       then 'n/a'
       else a_code
       end AS a_code
from   table_name
where  b_code = 'abc';

其中，对于示例数据：

CREATE TABLE table_name (a_code, b_code) AS
SELECT 'aaa', 'abc' FROM DUAL UNION ALL
SELECT 'aaa', 'abc' FROM DUAL;

输出：
| A_代码|
| - ------|
| 不适用|
| 不适用|
fiddle

一些样本数据和预期结果将有所帮助;同时看看这是否合理。
对于这样的示例数据，首先计算您感兴趣的行数，然后在final select中使用该值：

SQL> with table_name(a_code, b_code) as
  2    (select '1', 'abc' from dual union all
  3     select '2', 'abc' from dual union all
  4     select '3', 'def' from dual
  5    ),
  6  temp as
  7    (select count(*) cnt
  8     from table_name
  9     where b_code = 'abc'
 10    )
 11  select a.a_code, a.b_code, t.cnt,
 12    case when t.cnt > 1 then 'n/a' else a.a_code end result
 13  from table_name a cross join temp t;

A_CODE  B_CODE           CNT RESULT
------- ------- ------------ -------
1       abc                2 n/a
2       abc                2 n/a
3       def                2 n/a

SQL>

如果没有至少两个abc：

SQL> with table_name(a_code, b_code) as
  2    (select '1', 'abc' from dual union all
  3     select '2', 'abx' from dual union all   --> was 'abc' before
  4     select '3', 'def' from dual
  5    ),
  6  temp as
  7    (select count(*) cnt
  8     from table_name
  9     where b_code = 'abc'
 10    )
 11  select a.a_code, a.b_code, t.cnt,
 12    case when t.cnt > 1 then 'n/a' else a.a_code end result
 13  from table_name a cross join temp t;

A_CODE  B_CODE           CNT RESULT
------- ------- ------------ -------
3       def                1 3
2       abx                1 2
1       abc                1 1

SQL>