创建一个视图，然后用简单的插件查询它，而不是查询表。
例如：

create view simple_student as -- and then your query below
select DISTINCT
  Student_number,
  case when min(sr.racecd) = max(sr.racecd) 
       then min(sr.racecd)
       else 'Two or more races' 
  end as races
from students
group by ...

然后插件可以从simple_student而不是student导出数据。

可以使用分析函数：

SELECT student_number,
       CASE
       WHEN num_races = 1
       THEN racecd
       ELSE 'Two or more races'
       END as races
FROM   (
  SELECT student_number,
         racecd,
         ROW_NUMBER() OVER (PARTITION BY student_number ORDER BY racecd) AS rn,
         COUNT(DISTINCT racecd) OVER (PARTITION BY student_number) AS num_races
  FROM   students
)
WHERE  rn = 1;

你可以对每个种族使用left join，将STUDENT_NUMBER中的null值替换为0，然后对所有的种族进行加法运算。如果结果等于实际的学生人数，那么它只是一个种族，你可以使用Coalesce（）得到它。

WITH
    students (STUDENT_NUMBER, STUDENT_NAME, RACE) AS
        (
            Select 5234, 'Jack', 'White' From Dual Union All
            Select 4343, 'John', 'Black' From Dual Union All
            Select 4343, 'John', 'White' From Dual Union All
            Select 4323, 'Lee',  'Asian' From Dual 
        )

Select  DISTINCT s.STUDENT_NUMBER, 
        CASE WHEN Nvl(w.STUDENT_NUMBER, 0) + Nvl(b.STUDENT_NUMBER, 0) + Nvl(a.STUDENT_NUMBER, 0) = s.STUDENT_NUMBER
             THEN  COALESCE(w.RACE, b.RACE, a.RACE)
        ELSE 'Two or more races'
        END "RACES"
From students s
Left Join ( Select STUDENT_NUMBER, RACE 
            From students
            Where RACE = 'White' ) w ON(w.STUDENT_NUMBER = s.STUDENT_NUMBER)
Left Join ( Select STUDENT_NUMBER, RACE 
            From students
            Where RACE = 'Black' ) b ON(b.STUDENT_NUMBER = s.STUDENT_NUMBER)
Left Join ( Select STUDENT_NUMBER, RACE 
            From students
            Where RACE = 'Asian' ) a ON(a.STUDENT_NUMBER = s.STUDENT_NUMBER)

R e s u l t :
STUDENT_NUMBER RACES           
-------------- -----------------
          4323 Asian             
          4343 Two or more races 
          5234 White

没有聚合、没有分组、没有分析函数......只有左连接、Nvl（）-s和一些数学知识。