Question

这就是我的代码的样子：

SELECT p.productid AS 'Redni broj',
    p.code AS 'Sifra robe',
    p.unitname AS 'Jedinica mere',
    di.quantity AS 'Kolicina',
    di.price 'Stara cena',
    di.itemvalue 'Stari iznos',
    di.price2 AS 'Nova cena',
    (di.price2 * di.quantity) AS 'Novi iznos',
    (di.price2 * di.quantity - di.itemvalue) AS 'Difference',
    SUM(di.price2 * di.quantity - di.itemvalue > 0) AS 'Positive difference',
    SUM(di.price2 * di.quantity - di.itemvalue < 0) AS 'Negative Difference'
FROM product AS p
LEFT JOIN productcategorypr AS pc ON p.productid = pc.productid
LEFT JOIN documentitem AS di ON p.productid = di.productid
GROUP BY p.productid

我需要将Diffrence过滤为2列，所有差异均低于0，所有差异均高于0，差异为NewPrice - Old Price ... 谢谢你的建议...

Answer 1

此表达式不正确：

SUM(di.price2 * di.quantity - di.itemvalue > 0) AS 'Positive difference'

您需要将其替换为CASE表达式：

SUM(CASE WHEN di.price2 * di.quantity > di.itemvalue THEN di.price2 * di.quantity - di.itemvalue ELSE 0 END) AS 'Positive difference'

使用Negative difference表达式做同样的事情：

SUM(CASE WHEN di.price2 * di.quantity < di.itemvalue THEN di.price2 * di.quantity - di.itemvalue ELSE 0 END) AS 'Negative Difference'

Answer 2

SUM（案例时间（di.price2 * di.quantity - di.itemvalue）＆gt; 0那么（di.price2 * di.quantity - di.itemvalue）ELSE 0 END）AS'积极差异'

SUM（案例时间（di.price2 * di.quantity - di.itemvalue）＆lt; 0 THEN（di.price2 * di.quantity - di.itemvalue）ELSE 0 END）AS'Degative Difference'

我需要小于0且大于0的值的SUM

2 个答案: