获得产品的最便宜价格

Question

我有两张桌子：products 和 prices

products

• id (PK)
• 姓名

prices

• id (PK)
• product_id（FK > 产品）
• 价格
• 原价

每个产品可能有多个价格。我想要实现的是一个返回我all products on-sale with its cheapest price的查询。

• on-sale = 价格
• 如果产品不是on-sale，则不应包含在结果中
• 如果产品有多个符合on-sale 条件的价格，则只返回最便宜的价格。

结果表应该有这些列

• products.id
• products.name
• prices.id
• prices.price
• prices.originalPrice

通过我的尝试，我最终解决了这个问题：#1055 - Expression #3 of SELECT list is not in GROUP BY clause and contains nonaggregated column 'tbl.price' which is not functionally dependent on columns in GROUP BY clause; this is incompatible with sql_mode=only_full_group_by。 请注意，我无法更改配置。

MySQL 版本： 5.7.22

我在此处上传了包含示例数据的 SQL 导出： https://www.dropbox.com/s/6ucdv6592dum6n6/stackoverflow_export.sql?dl=0

Answer 1

select pro.name, MIN(pri.price) from products pro
inner join price pri on pri.product_id = pro.id 
where pri.price < pri.originalPrice 
group by pro.name

这是一个没有任何数据的镜头：p 可能需要稍作调整

Answer 2

试试这个：

SELECT * 
FROM `products` pro
JOIN price pri on pri.productId = pro.id
WHERE pri.price < pri.originalPrice
AND pri.price = 
( 
    SELECT min(p.price) 
    FROM price p 
    WHERE p.productId = pro.id AND p.price < p.originalPrice 
)

Answer 3

希望这对你有用

SELECT *,MIN(price)  FROM (
SELECT name, products.id,price
FROM products
INNER JOIN productItems
   ON products.id = productItems.productId
WHERE price < originalPrice
ORDER BY (price-originalPrice)
) as tbl GROUP BY id;

或

SELECT *,MIN(diff)  FROM (
SELECT name, products.id,price,(price-originalPrice) as "diff"
FROM products
INNER JOIN productItems
   ON products.id = productItems.productId
WHERE price < originalPrice
ORDER BY products.id,(price-originalPrice)
) as tbl GROUP BY id;

Answer 4

这适用于您提供的保管箱链接：http://sqlfiddle.com/#!9/a6306d/3

 select pro.name, MIN(pri.price) from products pro
    inner join price pri on pri.productId = pro.id 
    where pri.price < pri.originalPrice 
    group by pro.name