任务：找到最贵价格的文章的经销商。
SELECT article, dealer, price
FROM shop s1
WHERE price=(SELECT MAX(s2.price)
FROM shop s2
WHERE s1.article = s2.article);
+---------+--------+-------+
| article | dealer | price |
+---------+--------+-------+
| 0001 | B | 3.99 |
| 0002 | A | 10.99 |
| 0003 | C | 1.69 |
| 0004 | D | 19.95 |
+---------+--------+-------+
上述示例使用相关的子查询，可能无效（请参见第14.2.10.7节“相关子查询”）。解决问题的其他可能性是在FROM或LEFT JOIN语句中使用不相关的子查询。
不相关的子查询：
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;
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;
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，s2行的值将为NULL。参见第14.2.9.2节“加入语法”。