如何解决oracle sql中的单分组问题

Posted 2023-05-09

【中文标题】如何解决oracle sql中的单分组问题

【英文标题】：how to solve a single-grouping problem in oracle sql

【发布时间】：2020-01-31 03:21:47

【问题描述】：

我正在尝试使用 oracle 中提供的 EMP 表来练习透视，方法是选择 job 作为列，ename 作为值。 这是我目前得到的查询。

     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)

但是输出是 (B)，因为从 (A) 中可以看出，有两个工作（文员、销售员）的值均为 4。

问题） 有没有办法修改（C），所以它只会选择一个作业，尽管（A），因为我只想为所有作业创建足够的列以在旋转后适应？

希望我说清楚了。

* 完成查询 *

  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;

【参考方案1】：

如果您使用的是 oracle 12c 或更高版本，您可以使用FETCH 子句，如下所示：

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

如果您使用的是较低版本的 oracle，则可以使用 analytical function，如下所示：

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

干杯！！

【讨论】：

您的查询肯定有效，但我无法将其与上述内容结合起来，但我想这是我的工作。我应该指定我正在使用的版本。谢谢！