MYSQL 将左连接中的重复行与所有数据合并为一行答案

【问题标题】：MYSQL combine duplicate rows in left join into one row with all dataMYSQL 将左连接中的重复行与所有数据合并为一行
【发布时间】：2021-06-07 19:16:40
【问题描述】：

我正在尝试使用多个左连接将 SQL 查询格式化为可读的 CSV 文件。当我进行左连接时，它会返回相同 ID 的多行（因为它应该连接多个非唯一行），但这对于可读的 CSV 文件是不可接受的。

我想要的是将每个重复的行合并为一行，其中包含重复数据的多列。

当前连接结果示例：

SELECT *
FROM people
LEFT JOIN attributes on ID

ID | Name | Attributes
1  | Ken  | Tall
1  | Ken  | Slender
1  | Ken  | Blonde
2  | John | Short

想要的结果（导出为 CSV）：

ID | Name | Attribute 1 | Attribute 2 | Attribute 3
1    Ken    Tall          Slender       Blonde
2    John   Short

我也尝试按 ID 分组，但是当我这样做时，它只返回每个 ID 的属性之一，这也是不可接受的。

也许我没有在正确的地方寻找，但我似乎找不到任何功能来帮助我完成这项工作。

提前致谢！！！

【问题讨论】：

标签： mysql sql join pivot

【解决方案1】：

你也可以使用 GROUP_CONCAT ，不需要知道属性的最大数量，并且可以用于 csv 导入。

SELECT id,name,group_concat(attributes SEPARATOR ';') AS attributes FROM people GROUP BY id,name;

【讨论】：

【解决方案2】：

你需要一个条件聚合，比如

SELECT ID, Name, 
       MAX( CASE WHEN rn = 1 THEN attributes END ) AS attribute1, 
       MAX( CASE WHEN rn = 2 THEN attributes END ) AS attribute2,
       MAX( CASE WHEN rn = 3 THEN attributes END ) AS attribute3 
  FROM
    (
     SELECT p.ID, Name, attributes,
            ROW_NUMBER() OVER (PARTITION BY ID ORDER BY ID) AS rn
       FROM people AS p
       LEFT JOIN attributes AS a
         ON a.ID = p.attribute_ID
     ) AS pa
 GROUP BY ID, Name

其中attribute_ID 是people 表中的假定列，如果DB 版本为8.0，则使用窗口函数

另一种选择是使用动态透视，其中不需要知道每个人有多少不同的属性，也不需要为每个属性的聚合编写每个条件，例如

SET @sql = NULL;

SELECT GROUP_CONCAT(
           DISTINCT
              CONCAT(
                    'MAX(CASE WHEN rn =', rn,' THEN attributes END) AS attribute',rn
                    )
       )
  INTO @sql
  FROM 
  (
   SELECT DISTINCT ROW_NUMBER() OVER (PARTITION BY id) AS rn 
     FROM people 
    ORDER BY rn                 
  ) AS r;

SET @sql = CONCAT('SELECT ID, Name, ',@sql,
                   ' FROM
                      (
                       SELECT p.ID, Name, attributes,
                              ROW_NUMBER() OVER (PARTITION BY ID ORDER BY ID) AS rn
                         FROM people AS p
                         LEFT JOIN attributes AS a
                           ON a.ID = p.attribute_ID
                       ) AS pa
                    GROUP BY ID, Name'); 

PREPARE stmt FROM @sql;
EXECUTE stmt;
DEALLOCATE PREPARE stmt;

Demo

【讨论】：

【解决方案3】：

您还可以轻松地将 GROUP_CONCAT 与 SUBSTRING_INDEX 一起使用，例如：

SELECT p.*, t.name
    , SUBSTRING_INDEX(t.attribs, ',', 1) as attribute1
    , SUBSTRING_INDEX( SUBSTRING_INDEX(t.attribs, ',', 2), ',' , -1) as attribute2
    , SUBSTRING_INDEX( SUBSTRING_INDEX(t.attribs, ',', 3), ',' , -1) as attribute3
    , SUBSTRING_INDEX( SUBSTRING_INDEX(t.attribs, ',', 4), ',' , -1) as attribute4
FROM people p
LEFT JOIN ( 
SELECT uid, name , CONCAT(GROUP_CONCAT(attribute),',,,,') as attribs FROM attributes GROUP BY name
) as t ON t.uid = p.uid;

样本

MariaDB [bernd]> select * from people;
+----+------+
| id | uid  |
+----+------+
|  1 |    1 |
|  2 |    2 |
+----+------+
2 rows in set (0.00 sec)

MariaDB [bernd]> select * from attributes;
+----+------+------+-----------+
| id | uid  | name | attribute |
+----+------+------+-----------+
|  1 |    1 | Ken  | Tall      |
|  2 |    1 | Ken  | Slender   |
|  3 |    1 | Ken  | Blonde    |
|  4 |    2 | John | Short     |
+----+------+------+-----------+
4 rows in set (0.00 sec)

MariaDB [bernd]> SELECT p.*, t.name
    -> , SUBSTRING_INDEX(t.attribs, ',', 1) as attribute1
    -> , SUBSTRING_INDEX( SUBSTRING_INDEX(t.attribs, ',', 2), ',' , -1) as attribute2
    -> , SUBSTRING_INDEX( SUBSTRING_INDEX(t.attribs, ',', 3), ',' , -1) as attribute3
    -> , SUBSTRING_INDEX( SUBSTRING_INDEX(t.attribs, ',', 4), ',' , -1) as attribute4
    -> FROM people p
    -> LEFT JOIN ( 
    -> SELECT uid, name , CONCAT(GROUP_CONCAT(attribute),',,,,') as attribs FROM attributes GROUP BY name
    -> ) as t ON t.uid = p.uid;
+----+------+------+------------+------------+------------+------------+
| id | uid  | name | attribute1 | attribute2 | attribute3 | attribute4 |
+----+------+------+------------+------------+------------+------------+
|  1 |    1 | Ken  | Tall       | Slender    | Blonde     |            |
|  2 |    2 | John | Short      |            |            |            |
+----+------+------+------------+------------+------------+------------+
2 rows in set (0.01 sec)

MariaDB [bernd]>

【讨论】：

【解决方案4】：

也许这会有所帮助

SELECT *
FROM (SELECT ID, Name, Attributes,'Attribute ' || ROW_NUMBER () OVER (PARTITION BY Name ORDER BY ID) as columns 
   FROM   people) ppl
PIVOT Attributes FOR columns IN('Attribute 1','Attribute 2','Attribute 3'))
ORDER BY ID

【讨论】：