Mysql 连接6个不重复的表

我试图从6个不同的表中获取数据，但每次都会显示多个重复。我读了很多这方面的书，但经过多次实验后，我并没有找到答案。希望有人知道答案。这是我的桌子和我的一个实验

用户-id、电子邮件

教师-id、用户id、姓名、电话

教师班级-id，班级id，教师id

类-id，类

教师纪律-id、学科id、教师id

学科-id， 纪律

我试图获得以下观点：

Name                 Email               Telephone         Classes     Disciplines
---------------------------------------------------------------------------------
  First name       first@gmail.com      2559857544         Va,VB,VIa   Html,CSS

相反，我的表如下所示：

Name                 Email               Telephone         Classes     Disciplines
---------------------------------------------------------------------------------
  First name       first@gmail.com      2559857544           Va          Html
  First name       first@gmail.com      2559857544           Vb          Html
  First name       first@gmail.com      2559857544           VIa         Html
  First name       first@gmail.com      2559857544           Va          CSS
  First name       first@gmail.com      2559857544           Vb          CSS
  First name       first@gmail.com      2559857544           VIa         CSS

这是我的密码

$sql_users_students = "SELECT * FROM users 
INNER JOIN teachers ON users.id = teachers.user_id
INNER JOIN teacher_class ON teachers.user_id = teacher_class.teacher_id
INNER JOIN classes ON teacher_class.class_id = classes.id
LEFT JOIN teacher_discipline ON teachers.user_id = teacher_discipline.teacher_id
left JOIN disciplines ON teacher_discipline.discipline_id = disciplines.id";

 <tbody>
 <?php foreach($teachers as $key => $info){
  echo "<tr><td>".$info["name"]."</td><td>".$info["email"]."</td><td>".$info["telephone"]."</td><td>".$info["class"]."</td><td>".$info["discipline"]."</td></td><td><a href=''php/delete_student.php''>Изтриване</a></td></tr>";} ?>                 
  </tr>
 </tbody>

$sql\u users\u students=“SELECT*FROM users”
users.id=teachers.user\u id上的内部联接教师
在teachers.user\u id=teacher\u class.teacher\u id上内部加入teacher\u类
教师上的内部联接类。类\u id=classes.id
左键在teachers.user\u id=teacher\u diction.teacher\u id上加入教师纪律
左键连接教师纪律上的纪律。纪律\u id=纪律。id”；

使用

GROUP by

和

GROUP_CONCAT（）

将这些值连接在一起，如下所示：

SELECT 
  name,
  email,
  telephone,
  group_concat(distinct classes) as classes,
  group_concat(distinct disciplines) as disciplines
FROM users 
INNER JOIN teachers ON users.id = teachers.user_id
INNER JOIN teacher_class ON teachers.user_id = teacher_class.teacher_id
INNER JOIN classes ON teacher_class.class_id = classes.id
LEFT JOIN teacher_discipline ON teachers.user_id = teacher_discipline.teacher_id
left JOIN disciplines ON teacher_discipline.discipline_id = disciplines.id
group by name, email, telephone

请使用组_concat（函数）

---

考虑GROMPY-CONTATA。如果您从不使用GROPY-CONTAT，您将是很好的-尤其是在应用程序代码的上下文中。

SELECT users.name, users.email, teachers.telephone, 
group_concat(classes.class ', ') as classes, group_concat(disciplines.discipline ', ') 
as disciplines 
FROM users 
INNER JOIN teachers ON users.id = teachers.user_id
INNER JOIN teacher_class ON teachers.user_id = teacher_class.teacher_id
INNER JOIN classes ON teacher_class.class_id = classes.id
LEFT JOIN teacher_discipline ON teachers.user_id = teacher_discipline.teacher_id
left JOIN disciplines ON teacher_discipline.discipline_id = disciplines.id";