Question

我目前有这个查询，它按降序选择最高的“拾取次数”。我需要只过滤前10行/最高数字。我怎样才能做到这一点？

我尝试在底部添加'WHERE ROWNUM＆lt; = 10'，但无济于事。

SELECT customer.company_name, COUNT (item.pickup_reference) as "Number of Pickups" 
FROM customer
JOIN item ON (customer.reference_no=item.pickup_reference) 
GROUP BY customer.company_name, item.pickup_reference
ORDER BY COUNT (customer.company_name) DESC;

感谢您的帮助！

Answer 1

您需要对其进行子查询以使rownum正常工作。

SELECT *
FROM
(
SELECT customer.company_name, COUNT (item.pickup_reference) as "Number of Pickups" 
FROM customer
JOIN item ON (customer.reference_no=item.pickup_reference) 
GROUP BY customer.company_name, item.pickup_reference
ORDER BY COUNT (customer.company_name) DESC
) 
WHERE rownum <= 10

您也可以使用排名功能，但考虑到相对简单，我不确定是否愿意。

Answer 2

使用等级的解决方案是这样的：

select customer.company_name, COUNT (item.pickup_reference) from (
    select distinct customer.company_name, COUNT (item.pickup_reference) ,
    rank() over ( order by count(item.pickup_reference) desc) rnk 
    from customer
    JOIN item ON (customer.reference_no=item.pickup_reference) 
    group by customer.company_name, item.pickup_reference
    order by COUNT (customer.company_name) )
where rnk < 10

使用'rownum'得到最高结果并没有给出预期的结果，因为它获得了10个没有ordred的第一行，然后订购它们（请在安德鲁的回复评论中通知这个，我不要没有权利添加评论。

从SQL Count查询中选择最佳结果，包括表连接 - Oracle

2 个答案: