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

所以现在我想获得前两行（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

我收到了与上面相同的结果......

注意：结果只是示例，可能会有3个，1个或更多行具有最大值，下次

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：已更改为使用排名

Answer 2

当你说你想要前两行时，我假设你并不是说你总是想要前两行 - 而是你想要那些最大值为'reserved'的行。

你可以尝试：

    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中运行，我猜它也应该在PL / SQL中。

Answer 3

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;

获取最大值为3列的行

3 个答案: