Mysql 为什么在执行此SQL查询时会得到重复的列？

我尝试在mySQL中执行一组查询。在最后一个查询中，我试图选择course_id列，但有两列标题为course_id且值完全相同，因此我遇到一个错误，即

查询错误：错误：ER\U NON\U UNIQ\U错误：字段中的“课程id”列 名单模棱两可

我也不知道如何解决这个问题。我只想让它选择一个名为course\u id的列（不过，我认为这不是最佳实践）

这是我的链接，这是我提出的问题

SELECT ID, course_id FROM teaches;

SELECT * FROM classroom WHERE capacity > 50;

SELECT * FROM section;

SELECT DISTINCT course_id FROM (SELECT * FROM classroom WHERE capacity > 50) AS c INNER JOIN section AS s ON c.room_no = s.room_no;


SELECT * FROM teaches AS i INNER JOIN (SELECT DISTINCT course_id FROM (SELECT * FROM classroom WHERE capacity > 50) AS c INNER JOIN section AS s ON c.room_no = s.room_no) AS j ON i.course_id = j.course_id;

-- SELECT course_id FROM teaches AS i INNER JOIN (SELECT DISTINCT course_id FROM (SELECT building, room_no, capacity FROM classroom WHERE capacity > 50) AS c INNER JOIN section AS s ON c.room_no = s.room_no) AS j ON i.course_id = j.course_id;

---

如果您不需要重复的

课程id

，请使用

和
：

SELECT *
FROM teaches i INNER JOIN
     (SELECT DISTINCT course_id
      FROM (SELECT * FROM classroom WHERE capacity > 50) c INNER JOIN 
           section s
           ON c.room_no = s.room_no
     ) c
     USING (course_id);

如果您不需要重复的
课程id
，请使用
和
：

SELECT *
FROM teaches i INNER JOIN
     (SELECT DISTINCT course_id
      FROM (SELECT * FROM classroom WHERE capacity > 50) c INNER JOIN 
           section s
           ON c.room_no = s.room_no
     ) c
     USING (course_id);

您需要从哪个表中选择课程id，可以是i.course\u id或j.course\u id

SELECT i.course_id FROM teaches AS i INNER JOIN (SELECT DISTINCT course_id FROM (SELECT building, room_no, capacity FROM classroom WHERE capacity > 50) AS c INNER JOIN section AS s ON c.room_no = s.room_no) AS j ON i.course_id = j.course_id;

由于在内部联接中，它们中的任何一个都将显示相同的数据。
您需要从您想要的课程id的哪个表中选择，无论是i.course\u id还是j.course\u id

SELECT i.course_id FROM teaches AS i INNER JOIN (SELECT DISTINCT course_id FROM (SELECT building, room_no, capacity FROM classroom WHERE capacity > 50) AS c INNER JOIN section AS s ON c.room_no = s.room_no) AS j ON i.course_id = j.course_id;

因为它在内部连接中，它们中的任何一个都将显示相同的数据。
我尝试过，但现在我得到了这个错误
查询错误：错误：ER_派生的\u必须有\u别名：每个派生表必须有自己的别名
我尝试过，但现在我得到了这个错误
查询错误：ER_派生的\u必须有\u别名：每个派生表必须有自己的别名