### 3.6.4 保持某一列的分组最大值的行

*任务：对于每件商品，找到价格最贵的经销商或经销商。*

这个问题可以用这样的子查询来解决：

```
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 |
+---------+--------+-------+
```

前面的示例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”](correlated-subqueries.html)）。解决该问题的其他可能性是在`从`条款，一个`左连接`，或带有窗口函数的公用表表达式。

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;
```

`左连接`：

```
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;
```

这`左连接`工作的基础是，当`s1.price`处于最大值时，没有`s2.price`具有更大的价值，因此相应的`s2.文章`值为`空值`.看[第 13.2.10.2 节，“JOIN 子句”](join.html).

带窗函数的常用表表达式：

[](<>)[](<>)

```
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;
```
