Question

显示在雇用最多雇员人数的当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

星期日是1，星期一是2，......等......

现在我需要选择具有最大重复次数的那个。

通过查看表格，我们知道它是3（星期二）。我知道我需要使用子查询来获取它，它有吗？

Answer 1

我倾向于使用分析函数：

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;

Answer 2

一种方法，将查询扩展到上方（SQL Fiddle Example）：

SELECT FIRST_NAME, to_char("hire_date", 'd') AS DOW
FROM EMPLOYEES
WHERE to_char("hire_date", 'd') =
  (
      SELECT b.DOW 
      FROM
      (
        select a.*, ROWNUM rnum 
        from (
          SELECT to_char("hire_date", 'd') AS DOW, COUNT(1) AS cnt
          FROM EMPLOYEES 
          GROUP BY to_char("hire_date", 'd')
          ORDER BY cnt DESC
        ) a
        where rownum = 1
      ) b
  )

Answer 3

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')
 );

SQLFiddle

Oracle数据库sql查询。有？

3 个答案: