3.6.4保留某列的分组最大值的行

 

任务：对于每件商品，找到价格最昂贵的经销商。

这个问题可以通过像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，该查询可能效率不高（请参见第13.2.11.7节“相关的子查询”）。解决该问题的其他可能性是在FROM子句中使用不相关的子查询，a LEFT JOIN或带有窗口函数的公用表表达式。

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
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;

根据以下LEFT JOIN原理进行工作：当 s1.price处于最大值时，不 s2.price存在更大的值，因此对应的s2.article值为 NULL。请参见第13.2.10.2节“ JOIN子句”。

带有窗口功能的常用表表达式：

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;