第一次在这里提问，如果格式不正确，敬请谅解。

我正在尝试从多个表中提取一些信息以构建报告。所述表格来自 Magento 实例，以防有帮助。

所以，我有四个正在使用的表: customer_entity、customer_address_entity、customer_address_entity_text 和 customer_address_var_char

在customer_entity中，我需要获取三个字段:entity_id、email和group_id

在 customer_address_entity 中，我需要获取 parent_id(等于上面的 entity_id)和 entity_id(与上面的不同)。

在 customer_address_entity_text 中，我需要 entity_id、attribute_id 和值。

在 customer_address_entity_varchar 中，我需要值、entity_id 和 attribute_id。

我的问题主要在于最后一张表。我需要获取六个属性 ID 的值，但我希望将值作为多列(名字、姓氏、电子邮件等)而不是仅作为值返回。

这是我当前的代码:

SELECT
customer_entity.entity_id as "Customer ID", email, customer_address_entity_text.value as "Street Address", customer_address_entity_varchar.value
FROM
customer_entity, customer_address_entity, customer_address_entity_text, customer_address_entity_varchar
WHERE
customer_entity.group_id="2"
AND
customer_entity.entity_id = customer_address_entity.parent_id
AND
customer_address_entity.entity_id = customer_address_entity_text.entity_id
AND
customer_address_entity_text.attribute_id="24"
AND
customer_address_entity.entity_id = customer_address_entity_varchar.entity_id
AND
customer_address_entity_varchar.attribute_id in (19,21,25,26,27)

返回的结果格式如下:

Current

我想要的是:

Desired

是的，列顺序不是最好的，但我会在根据需要对列进行格式化后处理它。我尝试了几个子字符串查询(其中一个最终使服务器崩溃；哎呀)并使用了 EXISTS。是的，我的代码也很丑陋；我计划在获得所需格式的数据后清理它。

谢谢!

编辑:如果有人想要一份我最终部署的代码副本(向 Bernd 致敬):

SELECT
e.entity_id as "Customer ID", e.email, t.value AS "Street Address",
GROUP_CONCAT(IF(v.attribute_id = 19,v.value,NULL)) AS "First Name",
GROUP_CONCAT(IF(v.attribute_id = 21,v.value,NULL)) AS "Last Name",
GROUP_CONCAT(IF(v.attribute_id = 25,v.value,NULL)) AS "City",
GROUP_CONCAT(IF(v.attribute_id = 27,v.value,NULL)) AS "Region/State",
GROUP_CONCAT(IF(v.attribute_id = 26,v.value,NULL)) AS "Country"
FROM customer_entity e 
LEFT JOIN customer_address_entity a ON a.parent_id = e.entity_id
LEFT JOIN customer_address_entity_varchar v ON v.entity_id = a.entity_id
LEFT JOIN customer_address_entity_text t on t.entity_id = a.entity_id
WHERE e.group_id = 2
AND t.attribute_id = 24
GROUP BY v.entity_id

最佳答案

不必多次加入表。您可以对结果进行分组并获得值。这是一个示例

两个表

MariaDB [tmp]> select *from names;
+----+-------+----------------+
| id | name  | email          |
+----+-------+----------------+
|  1 | Bernd | bernd@bernd.de |
|  2 | David | david@david.de |
+----+-------+----------------+
2 rows in set (0.00 sec)

MariaDB [tmp]> select * from customer_address_entity;
+-----------+--------------+------------------+
| entity_id | attribute_id | value            |
+-----------+--------------+------------------+
|         1 |            1 | Duesseldorf      |
|         1 |            2 | 40211            |
|         1 |            3 | berlinerplatz 55 |
|         1 |            4 | 0211 / 1234567   |
|         2 |            1 | Bremen           |
|         2 |            2 | 21334            |
|         2 |            3 | Aachenerstr. 99  |
|         2 |            4 | 0432 / 7890111   |
+-----------+--------------+------------------+
8 rows in set (0.00 sec)

加入表

MariaDB [tmp]> SELECT
    ->   n.*,a.*
    -> FROM `names` n
    -> LEFT JOIN customer_address_entity a ON a.entity_id = n.id;
+----+-------+----------------+-----------+--------------+------------------+
| id | name  | email          | entity_id | attribute_id | value            |
+----+-------+----------------+-----------+--------------+------------------+
|  1 | Bernd | bernd@bernd.de |         1 |            1 | Duesseldorf      |
|  1 | Bernd | bernd@bernd.de |         1 |            2 | 40211            |
|  1 | Bernd | bernd@bernd.de |         1 |            3 | berlinerplatz 55 |
|  1 | Bernd | bernd@bernd.de |         1 |            4 | 0211 / 1234567   |
|  2 | David | david@david.de |         2 |            1 | Bremen           |
|  2 | David | david@david.de |         2 |            2 | 21334            |
|  2 | David | david@david.de |         2 |            3 | Aachenerstr. 99  |
|  2 | David | david@david.de |         2 |            4 | 0432 / 7890111   |
+----+-------+----------------+-----------+--------------+------------------+
8 rows in set (0.00 sec)

对其进行分组并获取字段

MariaDB [tmp]> SELECT
    ->   n.name, n.email,
    ->   GROUP_CONCAT( IF ( a.attribute_id = 1 , a.value,NULL)) AS city,
    ->   GROUP_CONCAT( IF ( a.attribute_id = 2 , a.value,NULL)) AS plz,
    ->   GROUP_CONCAT( IF ( a.attribute_id = 3 , a.value,NULL)) AS street,
    ->   GROUP_CONCAT( IF ( a.attribute_id = 4 , a.value,NULL)) AS phone
    -> FROM `names` n
    -> LEFT JOIN customer_address_entity a ON a.entity_id = n.id
    -> GROUP BY a.entity_id;
+-------+----------------+-------------+-------+------------------+----------------+
| name  | email          | city        | plz   | street           | phone          |
+-------+----------------+-------------+-------+------------------+----------------+
| Bernd | bernd@bernd.de | Duesseldorf | 40211 | berlinerplatz 55 | 0211 / 1234567 |
| David | david@david.de | Bremen      | 21334 | Aachenerstr. 99  | 0432 / 7890111 |
+-------+----------------+-------------+-------+------------------+----------------+
2 rows in set (0.00 sec)

MariaDB [tmp]>

关于php - mySQL:将一列的结果作为多列返回，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/32999059/