Question

我想显示仅剩一项的所有位置。使用SUM函数进行了尝试，但它仍然为我提供了多个具有相同位置的行，其中3个文章的总和为1

SELECT location, artnr, sum(qty) FROM location_stock GROUP BY location, artnr HAVING sum(qty) = 1

结果：

Location Art Nr Sum(Qty) 12345 5345 1 12345 5874 1 12345 4756 1

我想要的是具有一篇文章nr和一份数量的位置：

Location Art Nr Sum(Qty) 12345 5345 1

Answer 1

您可以使用row_number()窗口功能

   select * from ( SELECT location, artnr, sum(qty) as s,
    row_number() over(partition by location order by artnr asc) rn
    FROM location_stock
    GROUP BY location, artnr
    HAVING sum(qty) = 1
    ) a where a.rn=1

Answer 2

假设数量永远不会为负或为零，那么这应该起作用：

SELECT location, max(artnr) as artnr, sum(qty)
FROM location_stock
GROUP BY location
HAVING SUM(qty) = 1;

有关更通用的解决方案：

SELECT location, max(artnr) as artnr, sum(qty)
FROM location_stock
GROUP BY location
HAVING MIN(artnr) = MAX(artnr) AND
       SUM(qty) = 1;

如果您实际上并不关心数量，而只关心一个artnr，则删除SUM(qty)上的条件。

显示仅剩一项的位置

2 个答案: