显示只剩下一个项目的位置

Question

我想显示只剩下一个项目的所有位置。尝试使用SUM函数，但它仍然给我多个具有相同位置的行，3篇nr，其总和为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

我想要的是有一篇文章nr和一个数量的位置：

Location  Art Nr  Sum(Qty)
12345     5345    1

Answer 1

你可以使用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

假设数量从不为负或为零，那么这应该有效：

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

对于更一般的解决方案：

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

如果你实际上并不关心数量而只关心一个artnr，那么删除SUM(qty)上的条件。