Question

我在选择GROUP_CONCAT时遇到了问题，该行也应该包含与该问题类似的行号GROUP_CONCAT numbering 区别在于我必须按多列分组。

例如，我有2个表review和review_detail。
模式（MySQL v5.5）

create table review (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `submission_id` int(11) NOT NULL,
   PRIMARY KEY (`id`)
);

create table review_detail (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `review_id` int(11),
  `category_id` int(11),
  `rating` varchar(100),
  PRIMARY KEY (`id`)
);

insert into review (`id`, `submission_id`) values (1, 1), (2, 1), (3, 2), (4, 3), (5,1), (6,3), (7,2), (8,3);

insert into review_detail (`review_id`, `category_id`, `rating`)
values 
(1, 1, ' submission 1.1 cat 1'), (1, 2, ' submission 1.1 cat 2'),
(2, 1, ' submission 1.2 cat 1'), (2, 2, ' submission 1.2 cat 2'),
(3, 1, ' submission 2.1 cat 1'), (3, 2, ' submission 2.1 cat 2'),
(4, 1, ' submission 3.1 cat 1'), (4, 2, ' submission 3.1 cat 1'),
(5, 1, ' submission 1.3 cat 1'), (5, 2, ' submission 1.3 cat 2'),
(6, 1, ' submission 3.2 cat 1'), (6, 2, ' submission 3.2 cat 2'),
(7, 1, ' submission 2.2 cat 1'), (7, 2, ' submission 2.2 cat 2'),
(8, 1, ' submission 3.3 cat 1'), (6, 2, ' submission 3.3 cat 2')
;

查询＃1

SELECT * FROM review;

| id  | submission_id |
| --- | ------------- |
| 1   | 1             |
| 2   | 1             |
| 3   | 2             |
| 4   | 3             |
| 5   | 1             |
| 6   | 3             |
| 7   | 2             |
| 8   | 3             |

查询＃2

SELECT * FROM review_detail;

| id  | review_id | category_id | rating                |
| --- | --------- | ----------- | --------------------- |
| 1   | 1         | 1           |  submission 1.1 cat 1 |
| 2   | 1         | 2           |  submission 1.1 cat 2 |
| 3   | 2         | 1           |  submission 1.2 cat 1 |
| 4   | 2         | 2           |  submission 1.2 cat 2 |
| 5   | 3         | 1           |  submission 2.1 cat 1 |
| 6   | 3         | 2           |  submission 2.1 cat 2 |
| 7   | 4         | 1           |  submission 3.1 cat 1 |
| 8   | 4         | 2           |  submission 3.1 cat 1 |
| 9   | 5         | 1           |  submission 1.3 cat 1 |
| 10  | 5         | 2           |  submission 1.3 cat 2 |
| 11  | 6         | 1           |  submission 3.2 cat 1 |
| 12  | 6         | 2           |  submission 3.2 cat 2 |
| 13  | 7         | 1           |  submission 2.2 cat 1 |
| 14  | 7         | 2           |  submission 2.2 cat 2 |
| 15  | 8         | 1           |  submission 3.3 cat 1 |
| 16  | 6         | 2           |  submission 3.3 cat 2 |

每个提交的评论（外键= submission_id）都有多个带有category_id的review_detail条目（在我的示例中，只有2个类别（1,2）与查询无关）。 / p>

我必须创建一个选择，使我得到按submission_id和category_id分组的GROUP_CONCAT。

Concat字符串应返回
Reviewer 1: {rating}, Reviewer 2: {rating}, Reviewer 3: {rating} etc.。

例如对于submitt_id = 1和category_id = 1，组concat应该返回
Reviewer 1: submission 1.1 cat 1, Reviewer 2: submission 1.2 cat 1, Reviewer 3: submission 1.3 cat 1。

但是我无法正确设置concat组中的编号。

到目前为止，我已经进行了多次测试。

只有一个列计数器的组（有效）：
https://www.db-fiddle.com/f/6hA4Vft1mQGdw2Pew2An2T/3
Reviewer 1: submission 1.1 cat 1 of review 1 / Reviewer 2: submission 3.3 cat 1 of review 8 / Reviewer 3: submission 2.2 cat 1 of review 7 / Reviewer 4: submission 3.2 cat 1 of review 6 / ... etc.

SELECT
    --review.submission_id,
    review_detail.category_id,
    @i,
    GROUP_CONCAT(
        CONCAT(
            'Reviewer ',
            @i := @i + 1,
            ': ',
            rating,
            ' of review ',  review_id
        )
    SEPARATOR ' / '
    ) concatText,
    @i := 0
FROM
    review_detail
LEFT JOIN review ON review.id = review_detail.review_id,
    (
SELECT
    @i := 0
) init
GROUP BY
    review_detail.category_id
ORDER BY
    review_detail.category_id ASC
;

使用if和比较2个分组列的字符串进行测试（不起作用）：
https://www.db-fiddle.com/f/3woAVSw5hrav15jAmuWVdT/3
Reviewer 1: submission 1.1 cat 1 of review 1 / Reviewer 1: submission 1.2 cat 1 of review 2 / Reviewer 1: submission 1.3 cat 1 of review 5

SELECT
    submission_id,
    category_id,
    @i,
    @grp,
    CONCAT_WS("-", submission_id, category_id) AS catgroup,
    GROUP_CONCAT(
        CONCAT(
            'Reviewer ',
            @i := IF(
                @grp = CONCAT_WS("-", submission_id, category_id),
                @i + 1,
                IF(
                    @grp := CONCAT_WS("-", submission_id, category_id),
                    1,
                    1
                )
            ),
            ': ',
            rating,
            ' of review ',  review_id
        )
    ORDER BY review_id, submission_id, category_id 
    SEPARATOR ' / '
    ) concatText
FROM
    review_detail
LEFT JOIN review ON review.id = review_detail.review_id,
    (
SELECT
    @i := 0,
    @grp := ''
) init
GROUP BY
    review.submission_id,
    review_detail.category_id

那么有人对多列进行分组时，有谁知道正确地在GROUP_CONCAT调用中编号的方法吗？

Answer 1

您应避免在生产代码中使用用户定义的变量。

在manual for MySQL 5.6中说：

作为一般规则，除了在SET语句中，永远不要为用户变量分配一个值，并在同一变量中读取该值声明。

甚至在documentation for 8.0中也声明：

涉及用户变量的表达式的求值顺序为未定义。例如，不能保证SELECT @a, @a:=@a+1 首先评估@a，然后执行分配。

在将来的版本中，这可能完全不再起作用：

以前的MySQL版本可以为一个值分配一个值 SET以外的语句中的用户变量。此功能是 MySQL 8.0支持向后兼容，但受制于在将来的MySQL版本中删除。

所以这是一个没有用户定义变量的解决方案：

SELECT 
r.submission_id,
rd.category_id,
GROUP_CONCAT(CONCAT('Reviewer ', (SELECT COUNT(*) + 1 
                                  FROM review 
                                  JOIN review_detail ON review.id = review_detail.review_id 
                                  WHERE r.submission_id = review.submission_id 
                                  AND review_detail.category_id = rd.category_id 
                                  AND review_detail.id < rd.id
                                 ), ': ', rating, ' of review ', review_id) ORDER BY rating SEPARATOR ' / ') AS shorter_column_name
FROM 
review r 
JOIN review_detail rd ON rd.review_id = r.id
GROUP BY r.submission_id, rd.category_id;

返回

+---------------+-------------+-----------------------------------------------------------------------------------------------------------------------------------------------+
| submission_id | category_id | shorter_column_name                                                                                                                           |
+---------------+-------------+-----------------------------------------------------------------------------------------------------------------------------------------------+
|             1 |           1 | Reviewer 1:  submission 1.1 cat 1 of review 1 / Reviewer 2:  submission 1.2 cat 1 of review 2 / Reviewer 3:  submission 1.3 cat 1 of review 5 |
|             1 |           2 | Reviewer 1:  submission 1.1 cat 2 of review 1 / Reviewer 2:  submission 1.2 cat 2 of review 2 / Reviewer 3:  submission 1.3 cat 2 of review 5 |
|             2 |           1 | Reviewer 1:  submission 2.1 cat 1 of review 3 / Reviewer 2:  submission 2.2 cat 1 of review 7                                                 |
|             2 |           2 | Reviewer 1:  submission 2.1 cat 2 of review 3 / Reviewer 2:  submission 2.2 cat 2 of review 7                                                 |
|             3 |           1 | Reviewer 1:  submission 3.1 cat 1 of review 4 / Reviewer 2:  submission 3.2 cat 1 of review 6 / Reviewer 3:  submission 3.3 cat 1 of review 8 |
|             3 |           2 | Reviewer 1:  submission 3.1 cat 1 of review 4 / Reviewer 2:  submission 3.2 cat 2 of review 6 / Reviewer 3:  submission 3.3 cat 2 of review 6 |
+---------------+-------------+-----------------------------------------------------------------------------------------------------------------------------------------------+

Answer 2

修复您的查询。

基本问题是表本质上是未排序的，这就是MySQL优化程序删除ORDER BY的原因。

在MySQL中，将所有表放在FROM子句中足以使广告按顺序创建子查询，mysql会保留它。

在Mariadb中这已经足够了，您还添加了LIMIT 18446744073709551615，以便优化程序将其保留

模式（MySQL v5.5）

查询＃1

SELECT
    submission_id,
    category_id,
    @i,
    @grp,
    CONCAT_WS("-", submission_id, category_id) AS catgroup,
    GROUP_CONCAT(
        CONCAT(
            'Reviewer ',
            @i := IF(
                @grp = CONCAT_WS("-", submission_id, category_id),
                @i := @i + 1,
                IF(
                    @grp := CONCAT_WS("-", submission_id, category_id),
                    1,
                    1
                )
            ),
            ': ',
            rating,
            ' of review ',  review_id
        )
    ORDER BY review_id, submission_id, category_id 
    SEPARATOR ' / '
    ) concatText
FROM
    (SELECT review_id, submission_id, category_id,`rating` FROM review_detail
LEFT JOIN review ON review.id = review_detail.review_id
     ORDER BY review_id, submission_id, category_id ) t1,
    (
SELECT
    @i := 0,
    @grp := ''
) init


GROUP BY
    submission_id,
    category_id;

结果

| submission_id | category_id | @i  | @grp | catgroup | concatText                                                                                                                                    |
| ------------- | ----------- | --- | ---- | -------- | --------------------------------------------------------------------------------------------------------------------------------------------- |
| 1             | 1           | 0   |      | 1-1      | Reviewer 3:  submission 1.1 cat 1 of review 1 / Reviewer 2:  submission 1.2 cat 1 of review 2 / Reviewer 1:  submission 1.3 cat 1 of review 5 |
| 1             | 2           | 3   | 1-1  | 1-2      | Reviewer 3:  submission 1.1 cat 2 of review 1 / Reviewer 2:  submission 1.2 cat 2 of review 2 / Reviewer 1:  submission 1.3 cat 2 of review 5 |
| 2             | 1           | 3   | 1-2  | 2-1      | Reviewer 1:  submission 2.1 cat 1 of review 3 / Reviewer 2:  submission 2.2 cat 1 of review 7                                                 |
| 2             | 2           | 2   | 2-1  | 2-2      | Reviewer 2:  submission 2.1 cat 2 of review 3 / Reviewer 1:  submission 2.2 cat 2 of review 7                                                 |
| 3             | 1           | 2   | 2-2  | 3-1      | Reviewer 2:  submission 3.1 cat 1 of review 4 / Reviewer 1:  submission 3.2 cat 1 of review 6 / Reviewer 3:  submission 3.3 cat 1 of review 8 |
| 3             | 2           | 3   | 3-1  | 3-2      | Reviewer 3:  submission 3.1 cat 1 of review 4 / Reviewer 2:  submission 3.3 cat 2 of review 6 / Reviewer 1:  submission 3.2 cat 2 of review 6 |

View on DB Fiddle

Answer 3

您需要使用两步子查询来按审阅者编号排序。

SET @i := 0;
SET @grp := '';
SELECT
    submission_id,
    category_id,
    GROUP_CONCAT(
      CONCAT(
        'Reviewer ',
        i,
        ': ',
        rating,
        ' of review ',  review_id
      )
      ORDER BY i
      SEPARATOR ' / '
    ) concatText
FROM
-- second, add numbering
(
  SELECT *,
    @i := IF(
      @grp = @grp := CONCAT_WS('-',submission_id,category_id),
      @i + 1, 1) i
  FROM
  -- first, sort for numbering
  (
    SELECT
        review_id,
        submission_id,
        category_id,
        rating
    FROM review_detail LEFT JOIN review ON review.id = review_detail.review_id
    ORDER BY
        submission_id,
        category_id,
        review_id
  ) t1
) t2
GROUP BY
    submission_id,
    category_id
;

db fiddle

Answer 4

为完整起见，我还添加了解决方案，该解决方案如何在Mysql 8.0中完成

与COUNT（*）一起使用

with base as (
    
  SELECT
    review_id,
    submission_id,
    category_id,
    rating,
    count(*) over (partition by submission_id,category_id  order by review_id) num
  
    FROM review_detail LEFT JOIN review ON review.id = review_detail.review_id
    ORDER BY
        submission_id,
        category_id,
        review_id
)
select   
  submission_id,
         category_id,
         group_concat(concat('Reviewer', num, ': ', rating, ' of review ',  review_id ) separator ', ') concattext
from     base
group by 
submission_id,
category_id
;

OR ROW_NUMBER（）

with base as (
        SELECT
            review_id,
            submission_id,
            category_id,
            rating,
            ROW_NUMBER() over (partition by submission_id,category_id  order by review_id) num
        FROM review_detail 
        LEFT JOIN review ON review.id = review_detail.review_id
        ORDER BY
            submission_id,
            category_id,
            review_id
    )
    SELECT   
        submission_id,
        category_id,
        group_concat(concat('Reviewer', num, ': ', rating, ' of review ',  review_id  ) separator ', ') concattext
    from base
    group by 
        submission_id,
        category_id
;

DB Fiddle

GROUP_CONCAT具有多个列的分组依据

4 个答案: