在 PHP MySQL 上显示、连接、计数、分组 6 个表答案

【问题标题】：Show, Join, Count, Grouping 6 Tables on PHP MySQL在 PHP MySQL 上显示、连接、计数、分组 6 个表
【发布时间】：2017-02-01 09:14:10
【问题描述】：

我在使用 php 和 mysql 显示、连接和分组 6 个表时遇到问题。

我的桌子

loker_job：id_job、slug_job、职位、部门、级别、位置
loker_department: id_department, name_department
loker_level: id_level, name_level
loker_location：id_location，name_location
loker_applicant：id_job，文件
loker_user：id_user，姓名

我的查询

SELECT 
loker_job.slug_job,
loker_level.name_level,
loker_location.name_location,
loker_department.name_department,
loker_user.name,
loker_applicant.file,
count(*) as total_applicant
FROM
loker_job
LEFT JOIN loker_department ON loker_department.id_department =    loker_job.department
LEFT JOIN loker_level ON loker_level.id_level = loker_job.`level`
LEFT JOIN loker_location ON loker_location.id_location = loker_job.location
LEFT JOIN loker_user ON loker_user.id_user = loker_job.publisher
LEFT JOIN loker_applicant ON loker_applicant.id_job = loker_job.slug_job
GROUP by loker_job.slug_job

我想要这样的结果：

+----------+-----------------+-----------------+---------------+----------------------+-----------------+
| slug_job | name_level      | name_department | name_location | name_position        | total applicant |
+----------+-----------------+-----------------+---------------+----------------------+-----------------+
| 1111     | General Manager | Marketing       | America       | Vice General Manager | 3               |
| 2222     | E-officer       | Ecommerce       | Japan         | Officer              | 1               |
| 3333     | General Manager | Sales           | Rusia         | Vice General Manager | 0               |
| 4444     | General Manager | Marketing       | America       | Vice General Manager | 3               |
+----------+-----------------+-----------------+---------------+----------------------+-----------------+

结果

但是，我得到的结果是：

尽管从 db 中显示为 0 或什么都没有，但总有一行或多行结果为 1 个申请人
总行不正确，总数据中始终为 -1

任何帮助将不胜感激，谢谢

【问题讨论】：

标签： php mysql join count group-by

【解决方案1】：

“loker_user”表的第一个组和关系表的左外连接。并如下所述：

SELECT loker_job.slug_job, loker_level.name_level, loker_location.name_location,
loker_department.name_department, A.name, loker_applicant.file, IFNULL(A.total_applicant,0) as total_applicant
FROM loker_job
LEFT OUTER JOIN loker_department ON loker_department.id_department =   loker_job.department
LEFT OUTER JOIN loker_level ON loker_level.id_level = loker_job.`level`
LEFT OUTER JOIN loker_location ON loker_location.id_location = loker_job.location
LEFT OUTER JOIN loker_applicant ON loker_applicant.id_job = loker_job.slug_job
LEFT OUTER JOIN (
select id_user,name, count(*) as total_applicant from loker_user group by id_user, name
) as A ON A.id_user = loker_job.publisher

【讨论】：

它在“on 子句”中显示未知列“a.publisher”
将列名从“A.loker_job”更新为“loker_job.loker_job”。