Question

架构：

Cereal  
Cereal_name  
Buyer_name

Cereal_name是谷物的名称
Buyer_name是带来某种麦片的买家

我试图列出每种谷物及其购买次数，然后找到前4种谷物：答案的例子是：

'HoneyNut' 400  
'Cap'n Crunch' 400  
'Chocolate Flakes' 300  
'Frosted Flakes' 200

注意：有4个，因为其中一个具有相同的数字

我拥有的是：

Select Cereal_name, count(*) as numOfCount
From Cereal
Group By numOfCount Desc

我想使用rowNum<=3，但在这种情况下无法使用

提前致谢！

Answer 1

尝试：

SELECT * FROM
(SELECT 
    Cereal_name, 
    COUNT(*) as numOfCount
FROM Cereal
GROUP BY numOfCount
ORDER BY COUNT(*) DESC
) CEREALS
WHERE ROWNUM <= 3

我认为是甲骨文。

DB2：

可能是这个，但我不确定：

SELECT 
    Cereal_name, 
    COUNT(*) as numOfCount
FROM Cereal
GROUP BY numOfCount
ORDER BY COUNT(*) DESC
FETCH FIRST 3 ROWS ONLY

Answer 2

如果你不能使用DB2的兼容模式，那么使用标准的ANSI SQL窗口函数应该可以实现：

select *
from (
  select t1.*, 
         dense_rank() over (order by cereal_count) as rank
  from (
    select cereal_name, 
           count(*) as cereal_count
    from cereal
    group by cereal_name
  ) t1
) t2
where rank <= 3
order by rank desc

Answer 3

也许这个？

SELECT G.Cereal_name, G.numOfCount
FROM (
    Select Cereal_name, count(*) as numOfCount
    From Cereal
    Group By Cereal_name) G
inner join (
    Select DISTINCT count(*) as numOfCount
    From Cereal
    Group By Cereal_name
    order by numOfCount Desc
    FETCH FIRST 3 ROWS ONLY) C on C.numOfCount = G.numOfCount
order by G.numOfCount DESC

sql查询最高3

3 个答案: