Question

我一直在尝试将此查询写入1小时，但SQL Developer总是会抛出错误。

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);

00000 - ＆＃34;％s：无效的标识符＆＃34;

最后一行第20栏出错。

所以我找到了另一个解决方案，但又抛出了另一个错误：

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(*));

00000 - ＆＃34;群组功能嵌套太深＆＃34;

最后一行第25栏出错。

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

Answer 1

使用窗口功能：

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，您可以使用ROW_NUMBER()代替并获得一行，即使多个驱动程序具有相同的最大值。

Answer 2

试试这个。我按cnt按降序排序。然后选择它的顶行。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是否支持此功能，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);

无效的标识符错误，Oracle

4 个答案: