无效标识符错误，Oracle

Question

我一直在尝试编写这个查询 1 小时，但 SQL Developer 总是抛出错误。

SELECT d.driver_name, COUNT(*) AS cnt
  FROM Drivers d
  JOIN Fastest_laps fl ON d.ID_driver = fl.ID_driver
  GROUP BY d.driver_name
  HAVING cnt = MAX(cnt);

904. 00000 - "%s: 无效标识符"

错误在最后一行，第 20 列。

所以我想出了另一个解决方案，但抛出了另一个错误：

SELECT d.driver_name, COUNT(*) as cnt
  FROM Drivers d
  JOIN Fastest_laps fl ON d.ID_driver = fl.ID_driver
  GROUP BY d.driver_name
  HAVING COUNT(*) = MAX(COUNT(*));

935. 00000 - “组函数嵌套太深”

错误在最后一行，第 25 列。

编辑：谢谢gyus，你太棒了，几乎所有的回复都有效，但我必须选择一个......

Answer 1

使用窗口函数：

SELECT driver_name, cnt
FROM (SELECT d.driver_name, COUNT(*) AS cnt,
             MAX(COUNT(*)) OVER () as MAXcnt
      FROM Drivers d JOIN
           Fastest_laps fl
           ON d.ID_driver = fl.ID_driver
      GROUP BY d.driver_name
     ) d
WHERE cnt = MAXcnt;

您也可以使用RANK()或DENSE_RANK()来表达这一点：

SELECT driver_name, cnt
FROM (SELECT d.driver_name, COUNT(*) AS cnt,
             RANK() OVER (ORDER BY COUNT(*) DESC) as seqnum
      FROM Drivers d JOIN
           Fastest_laps fl
           ON d.ID_driver = fl.ID_driver
      GROUP BY d.driver_name
     ) d
WHERE seqnum = 1;

这种方法的优点是您可以改用ROW_NUMBER()并获得准确的一行，即使多个驱动程序具有相同的最大值。

Answer 2

尝试这个。 我按cnt降序排列。 然后从中选择顶行。 您可以将查询编辑为rownum <=2以获取前 2 行，依此类推。

           with tbl1 as
           (SELECT d.driver_name as driver_name, COUNT(*) AS cnt
            FROM Drivers d
            JOIN Fastest_laps fl ON d.ID_driver = fl.ID_driver
            GROUP BY d.driver_name
            order by cnt desc
            )
            select driver_name,cnt from tbl1
            where cnt = (select cnt from tbl1 rownum=1)

Answer 3

我不确定 Oracle 是否支持这一点，但请试一试：

SELECT d.driver_name, COUNT(*) as cnt
FROM Drivers d
  JOIN Fastest_laps fl ON d.ID_driver = fl.ID_driver
GROUP BY d.driver_name
ORDER BY cnt DESC
FETCH FIRST 1 ROW WITH TIES

或者使用公用表表达式：

with cte as
(
SELECT d.driver_name as driver_name, COUNT(*) AS cnt
FROM Drivers d
  JOIN Fastest_laps fl ON d.ID_driver = fl.ID_driver
GROUP BY d.driver_name
)
select driver_name, cnt
from cte
where cnt = (select max(cnt) from cte)

Answer 4

您必须将查询包装到内联视图中才能查询cnt ：

select *
from   (
         SELECT d.driver_name, COUNT(*) AS cnt
         FROM Drivers d
         JOIN Fastest_laps fl ON d.ID_driver = fl.ID_driver
         GROUP BY d.driver_name
       ) x
group
by     driver_name, cnt
having cnt = MAX(cnt);