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  • 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;
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  • 原文地址:https://www.cnblogs.com/owlin/p/13730978.html
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