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MYSQL - 按限制分组

发布于 2024-08-28 13:51:20 字数 4465 浏览 3 评论 0原文

有没有一种简单的方法可以将 GROUP BY 结果限制为前 2 个。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 <- 保持
波兰 1 15 5 0 <- 保留
波兰 23 12 6 3
波兰 2 5 0 0
波兰 3 0 5 0
波兰 4 0 0 5
爱尔兰 1 9 3 0 <- 保留
爱尔兰 2 3 0 0 <- 保留
爱尔兰 3 0 3 0
爱尔兰 4 0 0 3
法国 12 24 12 6 <- 保留
法国 1 3 1 0 <- 保留
法国 231 1 0 0
法国 2 1 0 0
法国 4 0 0 1
法国 3 0 1 0
谢谢

​ Jon

根据要求，我附上了表结构的副本和一些测试数据。我的目标是创建一个单一视图，其中包含每个唯一 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);
原文

Is there a simple way to LIMIT the GROUP BY results to the top 2. The following query returns all the results. Using 'LIMIT 2' reduces the overall list to the top 2 entries only.

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

Results in the following :-

poland  78 48 24 12   <- keep
poland   1 15  5  0   <- keep
poland  23 12  6  3
poland   2  5  0  0
poland   3  0  5  0
poland   4  0  0  5
ireland  1  9  3  0   <- keep
ireland  2  3  0  0   <- keep
ireland  3  0  3  0
ireland  4  0  0  3
france  12 24 12  6   <- keep
france   1  3  1  0   <- keep
france 231  1  0  0
france   2  1  0  0
france   4  0  0  1
france   3  0  1  0

Thanks
Jon

As requested I have attached a copy of the table structure and some test data. My goal is to create a single view that has the top 2 results from each unique rating_name

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);

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评论（3）

森末i 2024-09-04 13:51:20

我认为 MySQL 中没有简单的方法。一种方法是为按 rating_name 分组的每一行生成行号，然后仅选择 row_number 为 2 或更少的行。在大多数数据库中，您可以使用以下命令来完成此操作：

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：

france    1  2  0  0
france    2  1  0  0
ireland   1  4  2  0
ireland  21  3  1  0
poland    1  3  1  0
poland    2  1  0  0

I don't think that there is a simple way in MySQL. One way to do this is by generating a row number for each row partitioned in groups by rating_name, and then only select the rows with row_number 2 or less. In most databases you could do this using something like:

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

Unfortunately, MySQL doesn't support the ROW_NUMBER syntax. You can however simulate ROW_NUMBER using variables:

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

Result when run on your test data:

france    1  2  0  0
france    2  1  0  0
ireland   1  4  2  0
ireland  21  3  1  0
poland    1  3  1  0
poland    2  1  0  0
回复 原文
不如归去 2024-09-04 13:51:20

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”含义中的一些含糊之处。

查询如下：

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，如果同一个 rating_name 有多个 id_market 具有相同的“好”计数，那么您将获得两条以上的记录。例如，如果有三个爱尔兰 id_markets，其“良好”计数为 3（最高），那么如何显示前两个？你不能。因此查询将显示所有三个。

另外，如果有一个计数为“3”（最高）和两个计数为“2”，则无法显示前两个，因为您并列第二名，因此查询会显示所有三个。

如果您首先使用聚合结果集创建一个临时表，然后使用该临时表，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;

This is still possible via a single query, but it's a bit long, and there are some caveats, which I'll explain after the query. Though, they're not flaws in the query so much as some ambiguity in what "top two" means.

Here's the query:

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

The caveat is that if there is more than one id_market with the same "good" count for the same rating_name, then you will get more than two records. For example, if there are three ireland id_markets with a "good" count of 3, the highest, then how can you display the top two? You can't. So the query will show all three.

Also, if there were one count of "3", the highest, and two counts of "2", you couldn't show the top two, since you have a tie for second place, so the query shows all three.

The query will be simpler if you create a temporary table with the aggregate result set first, then work from that.

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;
回复 原文
风吹雨成花 2024-09-04 13:51:20
SUBSTRING_INDEX(
    GROUP_CONCAT(expr1 ORDER BY expr2 SEPARATOR ";"),
    ";",
    2  /* the GROUP_LIMIT */
)

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

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

expr1 can be like CONCAT(...). Involve REPLACE to hide any ";".

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