Question

我有一个复杂的查询要写（至少对我而言）。我有4张桌子 - ressources_department （ressource_record_id，account_id，department_id，有效） - ressources_type_users （ID，电子邮件，密码） - ressources_records （id，ressource_main_id，ressource_id，account_id） - ressources_records_details （ressource_record_id，ressource_field_id，ressource_data）

ressource_department持有特定部门和帐户的用户（ressources_records）。
ressource_type_users保存用户的基本信息（电子邮件，密码）及其唯一ID。
ressource_records保存用户唯一ID（ressource_main_id），也是用户详细信息的链接（ressources_records_details）
ressources_records_details保存用户数据。 Ressource_fields（int）是ressource_date的类型。防爆。：113是电子邮件地址的代码。

我必须编写一个查询来显示内联的所有信息。

到目前为止，我所做的是：

SELECT a.ressource_record_id, a.account_id, a.active, a.department_id, b.ressource_main_id, b.ressource_id 
FROM ressources_department a 
join ressources_records b on a.ressource_record_id = b.id
WHERE a.active=1

除了'ressources_records_details'数据之外，这几乎给了我所需要的东西。我可以通过php循环并查询每行的数据库，但我认为不是这样的。

我试过了：

SELECT a.ressource_record_id, a.account_id, a.active, a.department_id, b.ressource_main_id, b.ressource_id, max(if((c.ressource_type = 113),c.ressource_data,NULL)) AS email
FROM ressources_department a 
join ressources_records b on a.ressource_record_id = b.id
join ressources_records_details c on a.ressource_record_id = c.ressource_record_id 
group by c.ressource_record_id

但是有些记录丢失了...需要帮助，我被困住了。 THX！

更新：我得到了大量子查询我想要的结果，但这花费了很多服务器时间不是吗？

SELECT a.ressource_record_id, a.account_id, a.active, a.department_id, b.ressource_main_id, b.ressource_id,
(select ressource_data from ressources_records_details where ressource_record_id=a.ressource_record_id and ressource_type=113) as email,
(select ressource_data from ressources_records_details where ressource_record_id=a.ressource_record_id and ressource_type=101) as lastname,
(select ressource_data from ressources_records_details where ressource_record_id=a.ressource_record_id and ressource_type=100) as firstname,
(select ressource_data from ressources_records_details where ressource_record_id=a.ressource_record_id and ressource_type=114) as address,
(select ressource_data from ressources_records_details where ressource_record_id=a.ressource_record_id and ressource_type=115) as zip,
(select ressource_data from ressources_records_details where ressource_record_id=a.ressource_record_id and ressource_type=116) as city
FROM ressources_department a 
join ressources_records b on a.ressource_record_id = b.id

Answer 1

您可以使用CASE语句在没有子查询的情况下执行此操作：

SELECT rd.ressource_record_id, rd.account_id, rd.active, 
    rd.department_id, rr.ressource_main_id, rr.ressource_id,
CASE WHEN rrd.resource_type = 113 THEN ressource_data ELSE NULL END AS email,
CASE WHEN rrd.resource_type = 101 THEN ressource_data ELSE NULL END AS lastname,
CASE WHEN rrd.resource_type = 100 THEN ressource_data ELSE NULL END AS firstname,
CASE WHEN rrd.resource_type = 114 THEN ressource_data ELSE NULL END AS address,
CASE WHEN rrd.resource_type = 115 THEN ressource_data ELSE NULL END AS zip,
CASE WHEN rrd.resource_type = 116 THEN ressource_data ELSE NULL END AS city
FROM ressources_department rd 
JOIN ressources_records rr 
    ON rb.ressource_record_id = rr.id
JOIN ressources_records_details rrd 
    ON rrd.ressource_record_id = rd.ressource_record_id
GROUP BY rd.ressource_record_id, rd.account_id, 
    rd.active, rd.department_id, rr.ressource_main_id, rr.ressource_id

复杂的4个链接表MYSQL查询

1 个答案: