with music_id_t as ( # 找出id为1的人关注的人喜欢听的歌 select music_id from music_likes where user_id in ( # 找出userid为1的用户关注的人的id select follower_id from follow where user_id = 1 ) and music_id not in ( # 找出id为1的人喜欢的音乐 select music_id from music_likes where user_id = 1 ) ) select music_name from music # 内连接筛选出不属于...
