Question

是否有更方便，更简单的方法来编写以下SQL代码？

SELECT
              products.name,
              (SELECT COUNT(*) FROM `sales` WHERE sales.product_id = products.id AND sales.payment_method = 'debt') AS debt, 
              (SELECT COUNT(*) FROM `sales` WHERE sales.product_id = products.id AND sales.payment_method = 'cash') AS cash,
              (SELECT COUNT(*) FROM `sales` WHERE sales.product_id = products.id AND sales.payment_method = 'free') AS free

            FROM `sales`, `products`
            WHERE `sales`.`product_id` = `products`.`id`
            GROUP BY `products`.`id`

Answer 1

您可以根据/var/log/apache2/akeneo-pim_error.log：

进行条件SUM()聚合

Payment_method

我还添加了别名以使其更具可读性，并将Select P.name, Sum(Case When S.Payment_method = 'debt' Then 1 Else 0 End) As Debt, Sum(Case When S.Payment_method = 'cash' Then 1 Else 0 End) As Cash, Sum(Case When S.Payment_method = 'free' Then 1 Else 0 End) As Free From Sales S Join Products P On S.Product_id = P.Id Group By P.Name更改为GROUP BY而不是P.Name。 P.Id语句中没有显示P.Id，而SELECT允许这样做，结果会留给编译器，这可能会给您带来意想不到的结果。

Answer 2

SELECT  p.name,
        SUM(IF(s.payment_method = 'debt',1,0)) debt,
        SUM(IF(s.payment_method = 'cash',1,0)) cash,
        SUM(IF(s.payment_method = 'free',1,0)) free
FROM products p
JOIN sales s
ON  s.product_id = p.id
GROUP BY p.id

更新根据@jpw评论，我同意可以简化为：

SELECT  p.name,
        SUM(s.payment_method = 'debt') debt,
        SUM(s.payment_method = 'cash') cash,
        SUM(s.payment_method = 'free') free
FROM products p
JOIN sales s
ON  s.product_id = p.id
GROUP BY p.id

带有count的SQL嵌套子查询

2 个答案: