任务：对于每篇文章，找到价格最高的经销商。Task: For each article, find the dealer or dealers with the most expensive price.

这个问题可以通过如下子查询来解决：This problem can be solved with a subquery like this one:

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，这可能效率不高（参见第 15.2.15.7 节，“相关子查询”）。解决该问题的其他方法包括在FROM子句中使用非相关子查询、使用LEFT JOIN，或使用带窗口函数的公用表表达式。The preceding example uses a correlated subquery, which can be inefficient (see Section 15.2.15.7, “Correlated Subqueries”). Other possibilities for solving the problem are to use an uncorrelated subquery in the FROM clause, a LEFT JOIN, or a common table expression with a window function.

非相关子查询：Uncorrelated subquery:

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: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 处于最大值时，不存在更大的 s2.price，因此对应的 s2.article 的值为 NULL。参见 第 15.2.13.2 节，“JOIN 子句”。The LEFT JOIN works on the basis that when s1.price is at its maximum value, there is no s2.price with a greater value and thus the corresponding s2.article value is NULL. See Section 15.2.13.2, “JOIN Clause”.

使用窗口函数的公用表表达式：Common table expression with window function:

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;