我需要返回以下语句,但我只想返回每个销售值的 TOP 5 ......不是所有的记录。
Select ID, Code, sum(Sale) as Sale from TableName
Where Code = 11
Group By ID, code
我不想要这个!
Select TOP 5 ID, Code, sum(Sale) as Sale from TableName
Where Code = 11
Group By ID, code
我需要返回以下语句,但我只想返回每个销售值的 TOP 5 ......不是所有的记录。
Select ID, Code, sum(Sale) as Sale from TableName
Where Code = 11
Group By ID, code
我不想要这个!
Select TOP 5 ID, Code, sum(Sale) as Sale from TableName
Where Code = 11
Group By ID, code
With Cte as
( Select ID, Code, sale as Sales ,
row_number() over (partition by ID,code order by sale desc) as row_num
from TableName where code=11
)
Select Id,code,sum(sales) from cte
GROUP BY ID, code
WHERE row_num < 6
WITH TopSales AS (
SELECT *, RANK() OVER (PARTITION BY ID, Code ORDER BY Sale DESC) saleRank
FROM TableName
)
SELECT ID, Code, SUM(Sale) AS Sale
FROM TopSales
WHERE (Code = 11) AND (saleRank <= 5)
GROUP BY ID, code
可能你需要类似的东西:
;WITH sales (
SELECT
id,
code,
sale,
ROW_NUMBER() OVER (PARTITION BY id, code ORDER BY sales DESC) n
FROM
TableName
WHERE
Code = 11
)
SELECT
id, code, sum(sale) sale
FROM
sales
WHERE
n <= 5
GROUP BY
id,
code
ROW_NUMBER()
并PARTITION BY
帮助找到最后 5 次销售。然后你SUM
只顶(最高)5。
此查询返回每个(id、code)组的前 5 个销售额的总和。
select id, code, SUM (sale)
from
(
select id, code, sale,
ROW_NUMBER() over(partition by id, code order by sale desc) rn
from tablename
) v
where rn<=5
group by id, code
如果您只想返回每个组的前 5 个结果,您可以这样做:
with cte as
(ID, Code, Sale,ROW_NUMBER() over(partition by ID,
Code order by (select 0)) rownum
from TableName)
Select ID, Code, sum(Sale) as Sale from cte
Where Code = 11
and rownum<=5
Group By ID, code
如果您想返回每个组薪水最高的前 5 个结果,您可以这样做:
with cte as
(ID, Code, Sale,ROW_NUMBER() over(partition by ID,
Code order by Sale desc) rownum
from TableName)
Select ID, Code, sum(Sale) as Sale from cte
Where Code = 11
and rownum<=5
Group By ID, code
select id, code, sum(sale) as sale from tablename
where code = 11
group by id, code
order by sum(sale) desc
limit 5