任務：針對每篇文章，找出價格最貴的經銷商或多個經銷商。

這個問題可以使用子查詢來解決，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 |
+---------+--------+-------+

前面的範例使用了一個相關子查詢，這可能效率不高（請參閱15.2.15.7 節「相關子查詢」）。解決這個問題的其他可能性是在 FROM 子句中使用非相關子查詢、LEFT JOIN，或使用具有視窗函數的通用表格運算式。

非相關子查詢

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 的運作基礎是當 s1.price 達到最大值時，沒有 s2.price 的值比它大，因此對應的 s2.article 值為 NULL。請參閱15.2.13.2 節「JOIN 子句」。

具有視窗函數的通用表格運算式

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;