Question

第一次在这里问一个问题，如果最终被格式化得不好，请道歉。

我试图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，以防万一。

所以，我有四张桌子和我一起工作： customer_entity，customer_address_entity，customer_address_entity_text，和 customer_address_var_char

在 customer_entity 中，我需要获得三个字段：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（与上面不同）。

在 customer_address_entity_text 中，我需要entity_id，attribute_id和value。

在 customer_address_entity_varchar 中，我需要value，entity_id和attribute_id。

我的问题主要在于最后一张表。我需要为六个属性ID获取值，但我希望将值作为多个列（名字，姓氏，电子邮件等），而不是仅返回值。

这是我目前的代码：

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)

返回格式如下的结果：

Current

我想要的是：

Desired

是的，列顺序并不是最好的，但是在我按照需要格式化列后，我会处理它。我尝试了几个子字符串查询（其中一个最终崩溃了服务器; oops）并使用了EXISTS。是的，我的代码也很可怕;我计划以我正在寻找的格式查找数据后清理它。

谢谢！

编辑：是否有人想要我最终部署的代码副本（帽子提示给Bernd）：

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

Answer 1

没有必要多次加入表格。您可以对结果进行分组并获取值。这是一个样本

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)

加入表格

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)

将其分组并获取字段

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]>

Answer 2

MySQL支持连接表本身，因此您需要多次连接eav表并提取＆＃39; value＆＃39;领域：（实施例）

MyOtherType

mySQL：将一列作为多列返回结果

2 个答案: