MySQL 8.4 发行说明
任务: 查找每种商品价格最贵的经销商。
可以使用如下所示的子查询解决此问题
SELECT article, dealer, price
FROM shop s1
WHERE price=(SELECT MAX(s2.price)
FROM shop s2
WHERE s1.article = s2.article)
ORDER BY article;
+---------+--------+-------+
| article | dealer | price |
+---------+--------+-------+
| 0001 | B | 3.99 |
| 0002 | A | 10.99 |
| 0003 | C | 1.69 |
| 0004 | D | 19.95 |
+---------+--------+-------+
前面的示例使用了一个关联子查询,这可能会效率低下(请参阅 第 15.2.15.7 节“关联子查询”)。 解决该问题的其他可能性是在 FROM
子句中使用非关联子查询、LEFT JOIN
或带有窗口函数的公用表表达式。
非关联子查询
SELECT s1.article, dealer, s1.price
FROM shop s1
JOIN (
SELECT article, MAX(price) AS price
FROM shop
GROUP BY article) AS s2
ON s1.article = s2.article AND s1.price = s2.price
ORDER BY article;
LEFT JOIN
:
SELECT s1.article, s1.dealer, s1.price
FROM shop s1
LEFT JOIN shop s2 ON s1.article = s2.article AND s1.price < s2.price
WHERE s2.article IS NULL
ORDER BY s1.article;
LEFT JOIN
的工作原理是,当 s1.price
处于其最大值时,没有 s2.price
具有更大的值,因此相应的 s2.article
值为 NULL
。 请参阅 第 15.2.13.2 节“JOIN 子句”。
带有窗口函数的公用表表达式
WITH s1 AS (
SELECT article, dealer, price,
RANK() OVER (PARTITION BY article
ORDER BY price DESC
) AS `Rank`
FROM shop
)
SELECT article, dealer, price
FROM s1
WHERE `Rank` = 1
ORDER BY article;