Oracle数据库sql查询。有？

显示在雇员人数最多的一周中的某一天雇用的所有雇员的姓名

表:

Steven  06/17/1987
Neena   09/21/1989
Lex     01/13/1993
Alex    01/03/1990
Bruce   05/21/1991
Diana   02/07/1999
Kevin   11/16/1999
Trenna  10/17/1995
Curtis  01/29/1997
Randall 03/15/1998
Peter   07/09/1998
Eleni   01/29/2000
Ellen   05/11/1996
Jonath  03/24/1998
Kimber  05/24/1999
Jenni   09/17/1987
Michael 02/17/1996
Pat     08/17/1997
Shelley 06/07/1994
William 06/07/1994

到目前为止我所拥有的

SELECT FIRST_NAME, to_char(hire_date,'d') AS DOW FROM EMPLOYEES;

Steven  4
Neena   5
Lex     4
Alex    4
Bruce   3
Diana   1
Kevin   3
Trenna  3
Curtis  4
Randall 1
Peter   5
Eleni   7
Ellen   7
Jonath  3
Kimbe   2
Jenni   5
Michael 7
Pat     1
Shelley 3
William 3

星期天是1，星期一是2。。。等等

现在我需要选择一个最大重复次数的

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看看桌子，我们就知道现在是3点（星期二）。我知道我需要使用子查询来获取它，它是否具有？

单向，将您的查询扩展到上面（）：

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一种方法是将查询扩展到（）之上：

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我倾向于对此使用分析函数：

select e.*
from (SELECT to_char(hire_date, 'd') AS DOW, count(*) as cnt,
             row_number() over (order by count(*) desc) as seqnum
      FROM EMPLOYEES
     ) dow join
     EMPLOYEEs e
     on dow.DOW = to_char(e.hire_date, 'd') and seqnum = 1;

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我倾向于对此使用分析函数：

select e.*
from (SELECT to_char(hire_date, 'd') AS DOW, count(*) as cnt,
             row_number() over (order by count(*) desc) as seqnum
      FROM EMPLOYEES
     ) dow join
     EMPLOYEEs e
     on dow.DOW = to_char(e.hire_date, 'd') and seqnum = 1;

select * 
  from employees
 where to_char(hire_date, 'd') = (
     select max(to_char(hire_date, 'd')) keep (dense_rank last order by count(*))
       from employees 
      group by to_char(hire_date, 'd')
 );