как решить проблему одиночной группировки в oracle sql

Question

Я пытаюсь отработать сводку с таблицей EMP, предоставленной в oracle, выбрав задание в качестве столбца с ename в качестве значений. Это запрос, который я получил до сих пор.

     select distinct job, count(job) from emp group by job ... (A)
     -- Result)
     -- CLERK 4 
     -- SALESMAN 4
     -- PRESIDENT 1
     -- MANAGER 3
     -- ANALYST 2
    select t1.ename, t2.ename, t3.ename, t4.ename, t5.ename
    from
        (select ename, rank() over (order by ename) as n from emp where job = 'CLERK') t1, 
        (select ename, rank() over (order by ename) as n from emp where job = 'SALESMAN') t2, 
        (select ename, rank() over (order by ename) as n from emp where job = 'PRESIDENT') t3, 
        (select ename, rank() over (order by ename) as n from emp where job = 'MANAGER') t4, 
        (select ename, rank() over (order by ename) as n from emp where job = 'ANALYST') t5, 
        (select ename, rank() over (order by ename) as n from emp where job = 
        (select job ... (C)
        from emp 
        group by job
        having count(*) = (
            select max(count(*))
            from emp 
            group by job))
    ) t6 
    where t6.n = t1.n(+)
        and t6.n = t2.n(+)
        and t6.n = t3.n(+)
        and t6.n = t4.n(+)
        and t6.n = t5.n(+)
    order by t6.n;

    output: ORA-00937: not a single-group group function ... (B)

Но вывод (B), так как есть два задания (CLERK, SALESMAN), оба из которых имеют значение из 4, как вы можете видеть из (A).

Вопрос)
Есть ли способ изменить (C), чтобы он выбрал только одну работу, несмотря на (A), так как я хочу только создать достаточное количество столбцов для всех заданий после поворота?

Надеюсь, я все прояснил.

* Завершенный запрос *

  select t1.ename "Clerk", t2.ename "Salesman", t3.ename "President", t4.ename "Manager", t5.ename "Analyst"
            from
                (select ename, rank() over (order by ename) as n from emp where job = 'CLERK') t1, 
                (select ename, rank() over (order by ename) as n from emp where job = 'SALESMAN') t2, 
                (select ename, rank() over (order by ename) as n from emp where job = 'PRESIDENT') t3, 
                (select ename, rank() over (order by ename) as n from emp where job = 'MANAGER') t4, 
                (select ename, rank() over (order by ename) as n from emp where job = 'ANALYST') t5, 
                (select ename, rank() over (order by ename) as n from emp where job = 
                    (select job
                    from
                        (select *
                        from (
                            select job, count(job)
                            from emp
                            group by job
                            order by count(job) desc) 
                        where rownum = 1))) t6
                where t6.n = t1.n(+)
                and t6.n = t2.n(+)
                and t6.n = t3.n(+)
                and t6.n = t4.n(+)
                and t6.n = t5.n(+)
            order by t6.n;

Answer 1

Если вы используете oracle 12 c или выше, вы можете использовать предложение FETCH следующим образом:

select job, count(job) 
  from emp
group by job 
ORDER BY COUNT(JOB) DESC
FETCH FIRST ROW ONLY

Если вы используете более низкую версию oracle, вы можете использовать analytical function следующим образом:

Select job, cnt from
(select job, count(job) cnt, row_number() over (order by count(job) desc) as rn
  from emp
 group by job)
Where rn = 1

Ура !!