Task: For each article, find the dealer or dealers with the most expensive price.任务：针对每件商品，找到价格最贵的经销商。

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

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

The preceding example uses a correlated subquery, which can be inefficient (see Section 13.2.11.7, “Correlated Subqueries”). 前面的例子使用了一个相关的子查询，这可能是低效的（参见第13.2.11.7节，“相关子查询”）。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.解决这个问题的其他可能性是在FROM子句中使用不相关的子查询、LEFT JOIN或带有窗口函数的公共表表达式。

Uncorrelated subquery:不相关子查询：

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;

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. LEFT JOIN的工作原理是，当s1.price处于其最大值时，不存在具有更大值的s2.price，因此相应的s2.article值为NULL。See Section 13.2.10.2, “JOIN Clause”.请参阅第13.2.10.2节，“JOIN子句”。

Common table expression with window function:带有窗口函数的通用表表达式：

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;