sql - 获取具有 3 列的最大值的行

Question

我有这个 PL/SQL 查询：

SELECT customer_id, table_id, count(table_id) as reserved
FROM { derived table }
GROUP BY customer_id,table_id
ORDER BY reserved DESC

我有这个结果：

http://i.stack.imgur.com/ATfUw.png

所以现在我想获取前 2 行（根据reserved列的最大值），我尝试了这个查询：

SELECT customer_id,table_id,max(reserved) from
(
    SELECT customer_id, table_id, count(table_id) as reserved
    FROM { derived table }
    GROUP BY customer_id,table_id
    ORDER BY reserved DESC
)
GROUP BY customer_id, table_id

我收到与上述相同的结果...

注意：结果只是示例，下次可能会有 3、1 或更多行具有最大值

Answer 1

SELECT customer_id,table_id,reserved
FROM (SELECT customer_id,table_id, COUNT(*)as reserved, RANK() OVER (ORDER BY COUNT(*) DESC) AS ct_rank
      FROM { derived table }
      GROUP BY customer_id,table_id
      )sub
WHERE ct_rank = 1

编辑：更改为使用排名

Answer 2

您的查询非常接近。外部查询应该选择两行，而不是重新进行聚合：

SELECT customer_id, table_id, reserved from
(
    SELECT customer_id, table_id, count(table_id) as reserved
    GROUP BY customer_id,table_id
    ORDER BY reserved DESC
)
where rownum <= 2;

Answer 3

当您说您想要前 2 行时，我假设您并不是说您总是想要前 2 行，而是您想要具有最大值“保留”的行。

你可以试试：

    SELECT customer_id, table_id, count(table_id) as reserved
    FROM { derived table }
    GROUP BY customer_id, table_id
    HAVING count(table_id) = (
        SELECT top 1 count(table_id)
        FROM { derived table }
        GROUP BY customer_id, table_id
        ORDER BY reserved DESC
    ) 

我知道这可以在 T-SQL 中使用，我猜它也应该在 PL/SQL 中使用。