你可以使用exists逻辑来表达这一点：

SELECT user_id
FROM student_enrollment se1
WHERE course_id = 10 AND due_fee = 0 AND
      NOT EXISTS (
          SELECT 1
          FROM student_enrollment se2
          WHERE se2.user_id = se1.user_id AND
                se2.course_id <> 10
      );

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另一种方法，使用聚合：

SELECT user_id
FROM student_enrollment
WHERE due_fee = 0
GROUP BY user_id
HAVING SUM(course_id <> 10) = 0;

型

一个简单的方法是使用EXCEPT：查找所有参加课程#10的用户，除了那些还参加另一个课程的用户。

SELECT user_id FROM student_enrollment WHERE course_id = 10 and due_fee = 0
EXCEPT
SELECT user_id FROM student_enrollment WHERE course_id <> 10

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https://dev.mysql.com/doc/refman/8.0/en/except.html

使用条件聚合和筛选：

SELECT user_id
FROM
(
SELECT user_id, 
       count(case when course_id = 10 then 1 else 0 end) cnt_10,
       count(case when course_id <> 10 then 1 else 0 end) cnt_other
  FROM student_enrollment s
 GROUP BY user_id
) s
WHERE cnt_10>0 and cnt_other=0

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或

SELECT user_id
  FROM student_enrollment s
 GROUP BY user_id
HAVING count(case when course_id = 10 then 1 else 0 end)>0
   AND count(case when course_id <> 10 then 1 else 0 end)=0

型