MySQL每种产品的最低和最新价格

我有三张桌子

    price_history (Primary: id)
    +----+------------+------+---------+-------+------------+
    | id | id_product | sku  | id_shop | price |  date_add  |
    +----+------------+------+---------+-------+------------+

    | 1  | 11         | 101  | 1001    | 10    | 2017-07-01 |
    | 2  | 12         | 101  | 1002    | 15    | 2017-07-01 |
    | 3  | 13         | 101  | 1003    | 20    | 2017-07-01 |
    | 4  | 11         | 101  | 1001    | 11    | 2017-07-02 | <-- lowest latest

    | 5  | 14         | 102  | 1001    | 45    | 2017-07-01 |
    | 6  | 15         | 102  | 1002    | 45    | 2017-07-01 |
    | 7  | 16         | 102  | 1003    | 45    | 2017-07-01 | <-- shop 1003 is the lowest,
    latest among shops and should not be display, because the shop monitored is 1001
    | 8  | 15         | 102  | 1002    | 60    | 2017-07-02 | 
    | 9  | 14         | 102  | 1001    | 55    | 2017-07-02 | 

    |10  | 17         | 103  | 1001    | 90    | 2017-07-01 |
    |11  | 18         | 103  | 1002    | 90    | 2017-07-01 |
    |12  | 19         | 103  | 1003    | 90    | 2017-07-01 |
    |13  | 17         | 103  | 1001    | 100   | 2017-07-02 | <-- lowest latest 
    |14  | 18         | 103  | 1002    | 100   | 2017-07-02 |
    |15  | 19         | 103  | 1003    | 100   | 2017-07-02 |
    +----+------------+------+---------+-------+------------+

    product (primary: id_product)
    +------------+---- ---+-----+--------------+---------+
    | id_product | active | sku | product_name | id_shop |
    +------------+--------+-----+--------------+---------+
    | 11         | 1      | 101 | Red          | 1001    |
    | 12         | 1      | 101 | A bit red    | 1002    |
    | 13         | 1      | 101 | Very red0    | 1003    |
    | 14         | 1      | 102 | Blue         | 1001    |
    | 15         | 1      | 102 | A bit blue   | 1002    |
    | 16         | 1      | 102 | Very blue    | 1003    |
    | 17         | 1      | 103 | Green        | 1001    |
    | 18         | 1      | 103 | A bit green  | 1002    |
    | 19         | 1      | 103 | Very green   | 1003    |
    | 20         | 0      | 104 | Discontinued | 1001    |
    | 21         | 0      | 104 | Out of stock | 1002    |
    | 22         | 0      | 104 | Varnish      | 1003    |
    +------------+--------+-----+--------------+---------+

    shop (primary: id_shop)
    +---------+--------+
    | id_shop | name   |
    +---------+--------+
    | 1001    | Shop A |
    | 1002    | Shop B |
    | 1003    | SHop C |
    +---------+--------+

最后，我只需向受监控的店铺显示“店铺A”

预期结果：

    +----+------------+------+---------+-------+
    | id | id_product | sku  | id_shop | price | 
    +----+------------+------+---------+-------+
    | 4  | 11         | 101  | 1001    | 11    |
    |13  | 17         | 103  | 1001    | 100   |
    +----+------------+------+---------+-------+ 

sku 102的最低价格是1003店，所以我们不需要展示它们

JSFIDLE

多谢各位

---

需要帮助，仍未找到解决方案

这将为您提供商店每个sku的最大添加日期

SELECT sku, MAX(date_add)
FROM price_history
WHERE id_shop = 1001
GROUP BY sku;

然后，您可以将其包装到另一个查询中，并再次使用

price\u history

连接

，以获得结果，例如：

SELECT ph.id, ph.id_product, ph.sku, ph.id_shop, ph.price
FROM proce_history ph JOIN (
 SELECT sku, MAX(date_add) AS `date_add`
 FROM price_history
 WHERE id_shop = 1001
 GROUP BY sku
) a ON ph.sku = a.sku AND ph.date_add = a.date_add;

您只需使用带有limit子句的相关子查询即可实现这一点：

select *
from price_history h
where id_shop = 1001
and id = (select id
          from price_history p
          where h.sku = p.sku
          and h.id_shop = p.id_shop
          order by date_add desc, price asc
          limit 1
          );

以上是简单的解决方案，但对于较大的数据集，可能会遇到性能问题

然后，您可以使用以下基于联接的解决方案：

select *
from price_history
join (
    select sku, date_add, min(price) as price
    from price_history
    join (
        select sku, max(date_add) as date_add
        from price_history
        where id_shop = 1001
        group by sku
    ) t using (sku, date_add)
    where id_shop = 1001
    group by sku, date_add
) t using (sku, date_add, price)
where id_shop = 1001

或

展示你的预期结果too@JYoThI预期结果added@Gurwinder辛格：是的，你是对的<代码>日期\u添加

added@GurwinderSingh第二个符号是102，那么，在你的例子中，

11

是最新的，但不是最低的？哪一个是11？id\u product或id？@在表中需要时间戳列。这将解决问题。我有一个问题要问你，你在什么基础上说它是最低的，这个值是最新的。所以您需要时间戳列，然后是u最小时间戳和最大时间戳sku@karanarora

date\u add

added这可能是最简单的。我怀疑嵌套的不相关子查询会更快。@草莓-是。。我已经添加了基于联接和子查询的解决方案。@Gurvinder Singh更正了第二个价格\u历史sku 102

@inMILD该行怎么是最新的最低价格？不应该选择最新日期和最低价的行吗？我想我还是不完全理解你的逻辑。@Gurvindersingh因为1003店没有为sku 102增加新的价格，他是其他店中价格最低的，而他的价格是他自己最新的。
select *
from price_history
join (
    select sku, date_add, min(price) as price
    from price_history
    join (
        select sku, max(date_add) as date_add
        from price_history
        where id_shop = 1001
        group by sku
    ) t using (sku, date_add)
    where id_shop = 1001
    group by sku, date_add
) t using (sku, date_add, price)
where id_shop = 1001

select *
from price_history
join (
    select sku, date_add, id_shop, min(price) as price
    from price_history
    join (
        select sku, id_shop, max(date_add) as date_add
        from price_history
        where id_shop = 1001
        group by sku, id_shop
    ) t using (sku, date_add, id_shop)
    group by sku, date_add, id_shop
) t using (sku, date_add, price, id_shop);