Mysql 从第三个表中获取记录

我有三个表，还有重复的列名：）我想将相册连接到产品，将图像连接到相册。图像很多。尝试这样的查询，它会给我重复的产品。是否有机会在一个查询中获取所有信息

        SELECT
        *, p.name as nazwa, a.name as nazwa_al, i.name as obrazek
        FROM products p
        JOIN 
        albums a on p.album_id=a.id
        JOIN
          (SELECT *, images.name AS nazwa_im FROM images ORDER BY images.order ASC) i
        ON i.album_id=a.id
        ORDER BY p.order ASC

产品

+-------------+---------+------+-----+---------+----------------+
| Field       | Type    | Null | Key | Default | Extra          |
+-------------+---------+------+-----+---------+----------------+
| id          | int(11) | NO   | PRI | NULL    | auto_increment |
| name        | text    | NO   |     | NULL    |                |
| description | text    | NO   |     | NULL    |                |
| album_id    | int(11) | YES  |     | NULL    |                |
| order       | int(11) | NO   |     | NULL    |                |
+-------------+---------+------+-----+---------+----------------+

+-------+---------+------+-----+---------+----------------+
| Field | Type    | Null | Key | Default | Extra          |
+-------+---------+------+-----+---------+----------------+
| id    | int(11) | NO   | PRI | NULL    | auto_increment |
| name  | text    | NO   |     | NULL    |                |
+-------+---------+------+-----+---------+----------------+

+----------+---------+------+-----+---------+----------------+
| Field    | Type    | Null | Key | Default | Extra          |
+----------+---------+------+-----+---------+----------------+
| id       | int(11) | NO   | PRI | NULL    | auto_increment |
| name     | text    | NO   |     | NULL    |                |
| alt      | text    | NO   |     | NULL    |                |
| album_id | int(11) | NO   |     | NULL    |                |
| order    | int(11) | NO   |     | NULL    |                |
+----------+---------+------+-----+---------+----------------+

专辑

+-------------+---------+------+-----+---------+----------------+
| Field       | Type    | Null | Key | Default | Extra          |
+-------------+---------+------+-----+---------+----------------+
| id          | int(11) | NO   | PRI | NULL    | auto_increment |
| name        | text    | NO   |     | NULL    |                |
| description | text    | NO   |     | NULL    |                |
| album_id    | int(11) | YES  |     | NULL    |                |
| order       | int(11) | NO   |     | NULL    |                |
+-------------+---------+------+-----+---------+----------------+

+-------+---------+------+-----+---------+----------------+
| Field | Type    | Null | Key | Default | Extra          |
+-------+---------+------+-----+---------+----------------+
| id    | int(11) | NO   | PRI | NULL    | auto_increment |
| name  | text    | NO   |     | NULL    |                |
+-------+---------+------+-----+---------+----------------+

+----------+---------+------+-----+---------+----------------+
| Field    | Type    | Null | Key | Default | Extra          |
+----------+---------+------+-----+---------+----------------+
| id       | int(11) | NO   | PRI | NULL    | auto_increment |
| name     | text    | NO   |     | NULL    |                |
| alt      | text    | NO   |     | NULL    |                |
| album_id | int(11) | NO   |     | NULL    |                |
| order    | int(11) | NO   |     | NULL    |                |
+----------+---------+------+-----+---------+----------------+

图像

+-------------+---------+------+-----+---------+----------------+
| Field       | Type    | Null | Key | Default | Extra          |
+-------------+---------+------+-----+---------+----------------+
| id          | int(11) | NO   | PRI | NULL    | auto_increment |
| name        | text    | NO   |     | NULL    |                |
| description | text    | NO   |     | NULL    |                |
| album_id    | int(11) | YES  |     | NULL    |                |
| order       | int(11) | NO   |     | NULL    |                |
+-------------+---------+------+-----+---------+----------------+

+-------+---------+------+-----+---------+----------------+
| Field | Type    | Null | Key | Default | Extra          |
+-------+---------+------+-----+---------+----------------+
| id    | int(11) | NO   | PRI | NULL    | auto_increment |
| name  | text    | NO   |     | NULL    |                |
+-------+---------+------+-----+---------+----------------+

+----------+---------+------+-----+---------+----------------+
| Field    | Type    | Null | Key | Default | Extra          |
+----------+---------+------+-----+---------+----------------+
| id       | int(11) | NO   | PRI | NULL    | auto_increment |
| name     | text    | NO   |     | NULL    |                |
| alt      | text    | NO   |     | NULL    |                |
| album_id | int(11) | NO   |     | NULL    |                |
| order    | int(11) | NO   |     | NULL    |                |
+----------+---------+------+-----+---------+----------------+

---

为了简单起见，我不想修改db的结构。对我来说，最简单的解决方案是：一个产品=>一个相册=>多个图像

使用连接和别名来解决重复名称错误

您可以使用distint或group by，使结果按照相同的产品id对齐

SELECT
*, p.name as nazwa, a.name as nazwa_al, i.name as obrazek
FROM 
products p
JOIN 
albums a on p.album_id = a.id
JOIN
images i ON i.album_id = a.id
GROUP BY p.id
ORDER BY p.order ASC

如果右侧有多行，则需要使用

分组\u concat

SELECT
*, p.name as nazwa, a.name as nazwa_al, group_concat(i.name) as obrazek
FROM 
products p
JOIN 
albums a on p.album_id = a.id
JOIN
images i ON i.album_id = a.id
GROUP BY p.id
ORDER BY p.order ASC

---

非常感谢。它给了我语法错误，我必须在order by之前移动group by，现在它也只给了我一个图像“obrazek”-这就是I.name。未测试的查询。更新的查询。有帮助吗？还是需要帮助？@cssBlaster21895:检查我的最新答案？让我知道它是否有效？同一列上不能有两个

as

别名。所以要么

作为条形码

，要么

作为obrazek

，但不能两者兼而有之。@Barmar:copy-paste打字：（.Sorry.Updated-response。现在有任何错误吗？