任务：对每篇文章，找到最贵的经销商或经销商。

这个问题可以使用子查询来解决，如下所示：

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 |
+---------+--------+-------+

前面的示例使用了相关子查询，可以不效率（见第15.2.15.7节，“相关子查询”）。其他解决问题的可能性是使用无关联子查询在FROM子句中，一个LEFT JOIN，或者使用通用表达式和窗口函数。

无关联子查询：

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基于s1.price的最大值，不存在更大值的s2.price，因此对应的s2.article值为NULL。见第15.2.13.2节，“连接子句”。

通用表达式和窗口函数：

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;