MySQL Left Join + Min

Question

看似简单的MySQL问题，但我从来没有这么做过..

我有两张桌子，物品和价格，有一对多的关系。

Items Table
id, name

Prices Table
id, item_id, price

其中

prices.item_id = items.id

到目前为止我所拥有的：

SELECT items.id, items.name, MIN(prices.price)
FROM items
LEFT JOIN prices ON items.id = prices.item_id
GROUP BY items.id

如何以最低价格返回相应的price.id？谢谢！

Answer 1

如果有最多价格的多个价格记录，这将为Items中的记录返回多条记录：

select items.id, items.name, prices.price, prices.id
from items
left join prices on (
    items.id = prices.item_id 
    and prices.price = (
        select min(price)
        from prices
        where item_id = items.id
    )
);

Answer 2

基于MySQL 5.0参考手册中的最后一个示例 - 3.6.4. The Rows Holding the Group-wise Maximum of a Certain Column的新工作答案：

SELECT items.id, items.name, prices.price, prices.id
FROM items 
LEFT JOIN prices
    ON prices.item_id = items.id
LEFT JOIN prices AS filter
    ON filter.item_id = prices.item_id
    AND filter.price < prices.price
WHERE filter.id IS NULL

LEFT JOIN的工作原理是，当prices.price处于最小值时，没有filter.price值较小且filter行值为NULL

---

原始错误答案：

SELECT items.id, items.name, prices.price, prices.id
FROM items 
LEFT JOIN prices ON prices.item_id = items.id
ORDER BY prices.price ASC
LIMIT 1

Answer 3

好的，怎么样？

SELECT items.id, items.name, MIN(prices.price), prices.id
FROM items 
LEFT JOIN prices ON items.id = prices.item_id 
GROUP BY items.id, MIN(prices.price)  

Answer 4

SELECT top 1 items.id, items.name, prices.price, prices.id 
FROM items  
LEFT JOIN prices ON items.id = prices.item_id  
ORDER BY prices.price ASC