MYSQL 将左连接中的重复行与所有数据合并为一行

Question

我正在尝试使用多个左联接来格式化一个 SQL 查询到一个可读的 CSV 文件。当我进行左连接时，它 returns 同一 ID 的多行（因为它应该连接多个非唯一行），但是这对于可读的 CSV 文件来说是不可接受的。

我想要的是将每个重复的行合并为一行，多列用于重复数据。

当前加入结果示例：

SELECT *
FROM people
LEFT JOIN attributes on ID

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

期望的结果（导出为 CSV）：

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

我也尝试过按 ID 分组，但是当我这样做时，它只是 returns 每个 ID 的属性之一，这也是不可接受的。

也许我没有找对地方，但我似乎找不到任何功能来帮助我完成这个。

提前致谢！！！

Answer 1

您需要条件聚合，例如

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 是来自 people table 的假定列，如果数据库版本为 8.0[=16，则使用 window 函数=]

另一种选择是使用动态数据透视，其中不需要知道每个人有多少不同的属性，也不需要为属性的每个聚合编写每个条件，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

Answer 2

也可以使用GROUP_CONCAT，不需要知道最大属性数，csv导入也可以。

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

Answer 3

您也可以轻松地将 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]>

Answer 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

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