Question

我遇到一些查询问题。我需要从该查询中获取最大值和product_name：

select
    products.product_name,
    sum(product_invoice.product_amount) as total_amount
from 
    product_invoice
inner join 
    products on product_invoice.product_id = products.product_id
inner join 
    invoices on product_invoice.invoice_id = invoices.invoice_id
where 
    month(invoices.invoice_date) = 2
group by 
    products.product_name

此查询返回如下结果：

product_name  | total_amount
--------------+--------------
      chairs  |  70              
      ladders | 500  
      tables  | 150

如何摆脱这个问题：ladders 500？

Answer 1

您可以使用def isCollision(t1, t2): distance = math.sqrt(math.pow(t1.xcor()-t2.xcor(),2)+math.pow(t1.ycor()-t2.ycor(),2)) if distance < 15: return True else: return False和order by：

fetch first 1 row only

并非所有数据库都支持ANSI标准select p.product_name, sum(pi.product_amount) as total_amount from product_invoice pi inner join products p on pi.product_id = p.product_id inner join invoices i on pi.invoice_id = i.invoice_id where month(i.invoice_date) = 2 -- shouldn't you have the year here too? group by p.product_name order by total_amount fetch first 1 row only;子句。您可能需要使用fetch first，limit或其他一些构造。

请注意，我还引入了表别名 - 它们使查询更容易编写和读取。另外，如果您选择月份，那么您是否也应该选择年份？

在旧版本的SQL Server中，您将使用select top：

select top 1

要获取最高金额的所有行，请使用select top (1) p.product_name, sum(pi.product_amount) as total_amount from product_invoice pi inner join products p on pi.product_id = p.product_id inner join invoices i on pi.invoice_id = i.invoice_id where month(i.invoice_date) = 2 -- shouldn't you have the year here too? group by p.product_name order by total_amount;。

Answer 2

int a;
b=19;

Answer 3

如果您使用的是SQL Server，那么TOP可以提供解决方案：

SELECT TOP 1
    p.product_name,
    SUM(pi.product_amount) AS total_amount
FROM product_invoice pi
INNER JOIN products p
    ON pi.product_id = p.product_id
INNER JOIN invoices i
    ON pi.invoice_id = i.invoice_id
WHERE
    MONTH(i.invoice_date) = 2
GROUP BY
    p.product_name
ORDER BY
    SUM(pi.product_amount) DESC;

注意：如果可能有多个产品与最高金额相关联，并且您想要所有关联，那么请使用TOP 1 WITH TIES，例如

SELECT TOP 1 WITH TIES
... (the same query I have above)

从另一个查询中获取最大值

3 个答案: