MYSQL - 按限制分组

Question

有没有一种简单的方法可以将 GROUP BY 结果限制在前 2 位。以下查询返回所有结果。使用“LIMIT 2”将整个列表减少到仅前 2 个条目。

select distinct(rating_name), 
       id_markets, 
       sum(rating_good) 'good', 
       sum(rating_neutral)'neutral', 
       sum(rating_bad) 'bad' 
 from ratings 
 where rating_year=year(curdate()) and rating_week= week(curdate(),1)
 group by rating_name,id_markets
 order by rating_name, sum(rating_good) 
 desc

结果如下：-

波兰 78 48 24 12

谢谢 乔恩

---

根据要求，我附上了表格结构和一些测试数据的副本。我的目标是创建一个视图，其中包含来自每个唯一 rating_name 的前 2 个结果

CREATE TABLE `zzratings` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `id_markets` int(11) DEFAULT NULL,
  `id_account` int(11) DEFAULT NULL,
  `id_users` int(11) DEFAULT NULL,
  `dateTime` timestamp NULL DEFAULT CURRENT_TIMESTAMP,
  `rating_good` int(11) DEFAULT NULL,
  `rating_neutral` int(11) DEFAULT NULL,
  `rating_bad` int(11) DEFAULT NULL,
  `rating_name` varchar(32) DEFAULT NULL,
  `rating_year` smallint(4) DEFAULT NULL,
  `rating_week` tinyint(4) DEFAULT NULL,
  `cash_balance` decimal(9,6) DEFAULT NULL,
  `cash_spend` decimal(9,6) DEFAULT NULL,
  PRIMARY KEY (`id`),
  KEY `rating_year` (`rating_year`),
  KEY `rating_week` (`rating_week`),
  KEY `rating_name` (`rating_name`)
) ENGINE=MyISAM AUTO_INCREMENT=2166690 DEFAULT CHARSET=latin1;

INSERT INTO `zzratings` (`id`,`id_markets`,`id_account`,`id_users`,`dateTime`,`rating_good`,`rating_neutral`,`rating_bad`,`rating_name`,`rating_year`,`rating_week`,`cash_balance`,`cash_spend`)
VALUES
    (63741, 1, NULL, 100, NULL, 1, NULL, NULL, 'poland', 2010, 15, NULL, NULL),
    (63742, 1, NULL, 101, NULL, 1, NULL, NULL, 'poland', 2010, 15, NULL, NULL),
    (1, 2, NULL, 102, NULL, 1, NULL, NULL, 'poland', 2010, 15, NULL, NULL),
    (63743, 3, NULL, 103, NULL, NULL, 1, NULL, 'poland', 2010, 15, NULL, NULL),
    (63744, 4, NULL, 104, NULL, NULL, NULL, 1, 'poland', 2010, 15, NULL, NULL),
    (63745, 1, NULL, 105, NULL, 1, NULL, NULL, 'poland', 2010, 15, NULL, NULL),
    (63746, 1, NULL, 106, NULL, NULL, 1, NULL, 'poland', 2010, 15, NULL, NULL),
    (63747, 5, NULL, 100, NULL, 1, NULL, NULL, 'ireland', 2010, 15, NULL, NULL),
    (63748, 5, NULL, 101, NULL, 1, NULL, NULL, 'ireland', 2010, 15, NULL, NULL),
    (63749, 2, NULL, 102, NULL, 1, NULL, NULL, 'ireland', 2010, 15, NULL, NULL),
    (63750, 3, NULL, 103, NULL, NULL, 1, NULL, 'ireland', 2010, 15, NULL, NULL),
    (63751, 4, NULL, 104, NULL, NULL, NULL, 1, 'ireland', 2010, 15, NULL, NULL),
    (63752, 1, NULL, 105, NULL, 1, NULL, NULL, 'ireland', 2010, 15, NULL, NULL),
    (63753, 1, NULL, 106, NULL, NULL, 1, NULL, 'ireland', 2010, 15, NULL, NULL),
    (63754, 1, NULL, 100, NULL, 1, NULL, NULL, 'ireland', 2010, 15, NULL, NULL),
    (63755, 1, NULL, 101, NULL, 1, NULL, NULL, 'ireland', 2010, 15, NULL, NULL),
    (63756, 2, NULL, 102, NULL, 1, NULL, NULL, 'ireland', 2010, 15, NULL, NULL),
    (63757, 34, NULL, 103, NULL, NULL, 1, NULL, 'ireland', 2010, 15, NULL, NULL),
    (63758, 34, NULL, 104, NULL, NULL, NULL, 1, 'ireland', 2010, 15, NULL, NULL),
    (63759, 34, NULL, 105, NULL, 1, NULL, NULL, 'ireland', 2010, 15, NULL, NULL),
    (63760, 34, NULL, 106, NULL, NULL, 1, NULL, 'ireland', 2010, 15, NULL, NULL),
    (63761, 21, NULL, 100, NULL, 1, NULL, NULL, 'ireland', 2010, 15, NULL, NULL),
    (63762, 21, NULL, 101, NULL, 1, NULL, NULL, 'ireland', 2010, 15, NULL, NULL),
    (63763, 21, NULL, 102, NULL, 1, NULL, NULL, 'ireland', 2010, 15, NULL, NULL),
    (63764, 21, NULL, 103, NULL, NULL, 1, NULL, 'ireland', 2010, 15, NULL, NULL),
    (63765, 4, NULL, 104, NULL, NULL, NULL, 1, 'ireland', 2010, 15, NULL, NULL),
    (63766, 1, NULL, 105, NULL, 1, NULL, NULL, 'ireland', 2010, 15, NULL, NULL),
    (63767, 1, NULL, 106, NULL, NULL, 1, NULL, 'ireland', 2010, 15, NULL, NULL),
    (63768, 1, NULL, 100, NULL, 1, NULL, NULL, 'france', 2010, 15, NULL, NULL),
    (63769, 1, NULL, 101, NULL, 1, NULL, NULL, 'france', 2010, 15, NULL, NULL),
    (63770, 2, NULL, 102, NULL, 1, NULL, NULL, 'france', 2010, 15, NULL, NULL),
    (63771, 3, NULL, 103, NULL, NULL, 1, NULL, 'france', 2010, 15, NULL, NULL),
    (63772, 4, NULL, 104, NULL, NULL, NULL, 1, 'france', 2010, 15, NULL, NULL);

Answer 1

我认为 MySQL 中没有简单的方法。一种方法是为按 rating_name 分组的每一行生成一个行号，然后只选择 row_number 为 2 或更少的行。在大多数数据库中，您可以使用以下方式执行此操作：

SELECT * FROM (
    SELECT
        rating_name,
        etc...,
        ROW_NUMBER() OVER (PARTITION BY rating_name ORDER BY good) AS rn
    FROM your_table
) T1
WHERE rn <= 2

不幸的是，MySQL 不支持ROW_NUMBER 语法。但是，您可以使用变量模拟 ROW_NUMBER：

SELECT
    rating_name, id_markets, good, neutral, bad
FROM (
    SELECT
        *,
        @rn := CASE WHEN @prev_rating_name = rating_name THEN @rn + 1 ELSE 1 END AS rn,
        @prev_rating_name := rating_name
    FROM (
        SELECT
            rating_name,
            id_markets,
            SUM(COALESCE(rating_good, 0)) AS good,
            SUM(COALESCE(rating_neutral, 0)) AS neutral,
            SUM(COALESCE(rating_bad, 0)) AS bad
        FROM zzratings
        WHERE rating_year = YEAR(CURDATE()) AND rating_week = WEEK(CURDATE(), 1)
        GROUP BY rating_name, id_markets
    ) AS T1, (SELECT @prev_rating_name := '', @rn := 0) AS vars
    ORDER BY rating_name, good DESC
) AS T2
WHERE rn <= 2
ORDER BY rating_name, good DESC

在测试数据上运行时的结果：

法国 1 2 0 0 法国 2 1 0 0 爱尔兰 1 4 2 0 爱尔兰 21 3 1 0 波兰 1 3 1 0 波兰 2 1 0 0

Answer 2

这仍然可以通过单个查询实现，但它有点长，并且有一些注意事项，我将在查询后解释。不过，它们并不是查询中的缺陷，而是“前两个”的含义存在歧义。

这是查询：

SELECT ratings.* FROM
(SELECT rating_name, 
       id_markets, 
       sum(rating_good) 'good', 
       sum(rating_neutral)'neutral', 
       sum(rating_bad) 'bad' 
 FROM zzratings 
 WHERE rating_year=year(curdate()) AND rating_week = week(curdate(),1)
 GROUP BY rating_name,id_markets) AS ratings
LEFT JOIN
(SELECT rating_name, 
       id_markets, 
       sum(rating_good) 'good', 
       sum(rating_neutral)'neutral', 
       sum(rating_bad) 'bad' 
 FROM zzratings 
 WHERE rating_year=year(curdate()) AND rating_week= week(curdate(),1)
 GROUP BY rating_name,id_markets) AS ratings2
ON ratings2.good <= ratings.good AND
  ratings2.id_markets <> ratings.id_markets AND
  ratings2.rating_name = ratings.rating_name
LEFT JOIN
(SELECT rating_name, 
       id_markets, 
       sum(rating_good) 'good', 
       sum(rating_neutral)'neutral', 
       sum(rating_bad) 'bad' 
 FROM zzratings 
 WHERE rating_year=year(curdate()) AND rating_week= week(curdate(),1)
 GROUP BY rating_name,id_markets) AS ratings3
ON ratings3.good >= ratings2.good AND
  ratings3.id_markets <> ratings.id_markets AND
  ratings3.id_markets <> ratings2.id_markets AND
  ratings3.rating_name = ratings.rating_name
WHERE (ratings2.good IS NULL OR ratings3.good IS NULL) AND
  ratings.good IS NOT NULL
ORDER BY ratings.rating_name, ratings.good DESC

需要注意的是，如果对于同一个 rating_name 有多个具有相同“良好”计数的 id_market，那么您将获得两个以上的记录。例如，如果有三个爱尔兰 id_markets 的“好”计数为 3，最高的，那么如何显示前两个？你不能。所以查询将显示所有三个。

另外，如果有一个计数为“3”，最高，两个计数为“2”，则无法显示前两个，因为您与第二名并列，因此查询显示所有三个.

如果您首先使用聚合结果集创建一个临时表，然后从该表开始工作，则查询会更简单。

CREATE TEMPORARY TABLE temp_table
  SELECT rating_name, 
           id_markets, 
           sum(rating_good) 'good', 
           sum(rating_neutral)'neutral', 
           sum(rating_bad) 'bad' 
     FROM zzratings 
     WHERE rating_year=year(curdate()) AND rating_week= week(curdate(),1;

SELECT ratings.*
 FROM temp_table ratings
LEFT JOIN temp_table ratings2
ON ratings2.good <= ratings.good AND
  ratings2.id_markets <> ratings.id_markets AND
  ratings2.rating_name = ratings.rating_name
LEFT JOIN temp_table ratings3
ON ratings3.good >= ratings2.good AND
  ratings3.id_markets <> ratings.id_markets AND
  ratings3.id_markets <> ratings2.id_markets AND
  ratings3.rating_name = ratings.rating_name
WHERE (ratings2.good IS NULL OR ratings3.good IS NULL) AND
  ratings.good IS NOT NULL
ORDER BY ratings.rating_name, ratings.good DESC;

Answer 3

SUBSTRING_INDEX(
    GROUP_CONCAT(expr1 ORDER BY expr2 SEPARATOR ";"),
    ";",
    2  /* the GROUP_LIMIT */
)

expr1 可以类似于 CONCAT(...)。涉及 REPLACE 以隐藏任何“;”。