在加入最大（日期）MYSQL后使用最小值

我有三张桌子制造商，产品和价格

我想得到产品的最后一个价格，然后选择它们的最低价格

表制造商：

# manufacturers
id        name
 1        Manufacturer 1
 2        Manufacturer 2

表产品：

# products
id        name
 1        Product 1
 2        Product 2

表格价格：

# prices
id        price        manufacturerId        createdAt    
 1           10                     1        '2019-09-09 00:00:00'
 2           20                     1        '2019-09-10 00:00:00'
 3           11                     2        '2019-09-09 00:00:00'
 4           21                     2        '2019-09-10 00:00:00'

完整代码：

DROP DATABASE if exists ssg ;
CREATE DATABASE ssg;
USE ssg;

# Create database manufacturers
CREATE TABLE manufacturers (id INT(11) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
                            name VARCHAR(256) NOT NULL);
# Insert value
INSERT INTO manufacturers (name) VALUES ('Manufacturer 1');
INSERT INTO manufacturers (name) VALUES ('Manufacturer 2');


# Create database products
CREATE TABLE products (id INT(11) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
                       name VARCHAR(256) NOT NULL);
# Insert value
INSERT INTO products (name) VALUES ('Product 1');

# Create database prices
CREATE TABLE prices (id INT(11) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
                     productId INT(11) UNSIGNED NOT NULL,
                     price BIGINT UNSIGNED NOT NULL,
                     manufacturerId INT(11) UNSIGNED NOT NULL,
                     createdAt DATETIME NOT NULL);
# Insert value
INSERT INTO prices (productId, price, manufacturerId, createdAt) VALUES (1, 10, 1, '2019-09-09 00:00:00');
INSERT INTO prices (productId, price, manufacturerId, createdAt) VALUES (1, 20, 1, '2019-09-10 00:00:00');
INSERT INTO prices (productId, price, manufacturerId, createdAt)VALUES (1, 11, 2, '2019-09-09 00:00:00');
INSERT INTO prices (productId, price, manufacturerId, createdAt)VALUES (1, 21, 2, '2019-09-10 00:00:00');

# Query
SELECT products.id, products.name, lastValue.price as latestPrice, lastValue.manufacturerId
FROM products
LEFT JOIN(
        SELECT productId, COUNT(DISTINCT manufacturerId) AS total
        FROM prices
        GROUP BY prices.productId) counts ON counts.productId = products.id
        LEFT JOIN (
            SELECT prices.*
            FROM (
                    SELECT productId, MAX(createdAt) createdAt
                    FROM prices
                    GROUP BY productId) latest
                    JOIN prices ON latest.productId = prices.productId
                    AND prices.createdAt = latest.createdAt
            ) lastValue
        ON lastValue.productId = products.id

我得到：

id        name        latestPrice        manufacturerId
 1        Product 1            20                     1
 1        Product 1            21                     2

那么，我怎样才能收到只有最低最新价格的产品呢

我得把它寄进去。请先“构建架构”，然后“运行SQL”

---

对不起，我的英语不好。

在MySQL 8.0中，您只能通过窗口功能来实现这一点：

select id, name, price, manufacturerId
from (
    select 
        t.*,
        rank() over(order by price) rn2
    from (
        select
            p.id,
            p.name,
            i.price,
            i.manufacturerId,
            rank() over(partition by p.id order by i.createdAt desc) rn1
        from products p
        inner join prices i on i.productId = p.id
    ) t
    where rn1 = 1
) t
where rn2 = 1

这些短语包括：

• 首先按降序日期对每个产品的价格进行排名，然后根据每个产品的最新价格进行筛选
• 然后按升价对所有最新价格进行排序，并筛选其中最低的价格

：

id | name | price | manufacturerId -: | :-------- | ----: | -------------: 1 | Product 1 | 20 | 1 id |名称|价格|制造商id -: | :-------- | ----: | -------------: 1 |产品1 | 20 | 1

---

您使用的是哪个版本的MySQL？我使用的是MySQL 8.0.18

prices。样本数据也应该填充product_id

列。欢迎@Trọ恩古伊ễ南昌！这是一个问得很好的问题，因此我们可以利用更多的信息。