Php 正确使用notin子句

这里的场景是：

我有四张桌子

课程

学生

部门

注册课程

当学生要注册课程时，课程列表应该在课程表的下拉列表中。但已在“注册课程”表中的课程id不应加载到该课程下拉列表中。因为一个学生不能注册一门课2次。这里我使用的是

notin

操作，它将检索不在注册课程表中的课程id列表

但我在查询中遇到以下错误：

您的SQL语法有错误；检查手册 对应于要使用的正确语法的MariaDB服务器版本 第6行“注册课程。课程id限制0,25”附近

以下是查询：

SELECT students.name, students.email,departments.name as d_name,
        courses.name as c_name
        FROM students JOIN departments on students.department_id=departments.id 
        JOIN courses on departments.id = courses.department_id
        LEFT JOIN enroll_courses on enroll_courses.course_id=courses.id 
        WHERE courses.id NOT IN (enroll_courses.course_id) AND students.id=8

有谁能帮我找到解决办法吗？

试试这个

SELECT  students.name ,
        students.email ,
        departments.name AS d_name ,
        courses.name AS c_name
FROM    students
        JOIN departments ON students.department_id = departments.id
        JOIN courses ON departments.id = courses.department_id
        JOIN enroll_courses ON enroll_courses.course_id = courses.id
WHERE   students.id = 8
        AND courses.id NOT IN ( enroll_courses.course_id );

试着这样,

SELECT  students.name ,
        students.email ,
        departments.name AS d_name ,
        courses.name AS c_name
FROM    students
        JOIN departments ON students.department_id = departments.id
        JOIN courses ON departments.id = courses.department_id
        JOIN enroll_courses ON enroll_courses.course_id = courses.id
WHERE   students.id = 8
        AND courses.id NOT IN ( enroll_courses.course_id );

---

有几种方法可以获得所需的结果，这与原始查询非常接近：

SELECT  students.name ,
        students.email ,
        departments.name AS d_name ,
        courses.name AS c_name
FROM    students
JOIN departments
  ON students.department_id = departments.id
JOIN courses
  ON departments.id = courses.department_id
WHERE students.id = 8   
-- up to here it's all possible courses

-- now remove already enrolled courses
  AND courses.id NOT IN  
   (
     SELECT course_id
     FROM enroll_courses
     WHERE student_id = 8
   )

或相关版本：

  AND courses.id NOT IN  
   (
     SELECT course_id
     FROM enroll_courses
     WHERE student_id = students.student_id
   )

或不存在：

  AND NOT EXISTS 
   (
     SELECT *
     FROM enroll_courses
     WHERE student_id = students.student_id
       AND course_id = courses.course_id
   )

---

有几种方法可以获得所需的结果，这与原始查询非常接近：

SELECT  students.name ,
        students.email ,
        departments.name AS d_name ,
        courses.name AS c_name
FROM    students
JOIN departments
  ON students.department_id = departments.id
JOIN courses
  ON departments.id = courses.department_id
WHERE students.id = 8   
-- up to here it's all possible courses

-- now remove already enrolled courses
  AND courses.id NOT IN  
   (
     SELECT course_id
     FROM enroll_courses
     WHERE student_id = 8
   )

或相关版本：

  AND courses.id NOT IN  
   (
     SELECT course_id
     FROM enroll_courses
     WHERE student_id = students.student_id
   )

或不存在：

  AND NOT EXISTS 
   (
     SELECT *
     FROM enroll_courses
     WHERE student_id = students.student_id
       AND course_id = courses.course_id
   )

---

NOT IN（enroll\u courses.course\u id）应该是这样添加一些示例表数据和预期结果。也许您希望

不在（子查询）

中。

注册课程表的DDL是多少？应该有两列，
课程id
和
学生id
。是的，你是对的。有两列，第一列。学生证2。课程id不在（enroll\u courses.Course\u id）应该是这样添加一些示例表数据和预期结果。也许您希望
不在（子查询）
中。
注册课程表的DDL是多少？应该有两列，
课程id
和
学生id
。是的，你是对的。有两列，第一列。学生证2。课程id这应该如何工作，您加入
注册课程。课程id=courses.id
，然后检查
课程。id不在（enroll\u courses.course\u id
，两者都相同又不相同？我没有从您建议的上述查询中获得任何数据。这是怎么回事，您加入
enroll\u courses.course\u id=courses.id
，然后检查
courses.id不在（注册课程。课程id
，两者都相同又不相同？我没有从您建议的上述查询中获得任何数据。