user1 = User(username='john')
user2 = User(username='jim')
user1.intersect(user2).all()在使用MariaDB而非SQLite时引发错误。
这是错误：
sqlalchemy.exc.ProgrammingError：（_mysql_exceptions.ProgrammingError）（1064，“您的SQL语法有错误；请查看与您的MySQL服务器版本相对应的手册，以获取在'INTERSECT SELECT user.id附近使用的正确语法AS user_id，user.username AS user_username，user.email'位于第4行”）[SQL：'SELECT anon_1.user_id AS anon_1_user_id，anon_1.user_username AS anon_1_user_username，anon_1.user_email AS anon_1_user_email，anon_1.user_password Anon_1_user_password AS anon_1_user_first_name，anon_1.user_last_name AS anon_1_user_last_name，anon_1.user_about_me AS anon_1_user_about_me，anon_1.user_last_seen AS anon_1_user_last_seen，anon_1.user_member_since AS anon_1_user_member_since，anon_1.user_last_message_read_time AS anon_1_user_last_message_read_time，anon_1.user_last_request_read_time AS anon_1_user_last_request_read_time \ n从（SELECT user.id AS USER_ID，用户。用户名AS user_username，user.email AS user_email，user.password AS user_pa ssword，user.first_name AS user_first_name，user.last_name AS user_last_name，user.about_me AS user_about_me，user.last_seen AS user_last_seen，user.member_since AS user_member_since，user.last_message_read_time AS user_last_message_read_re_time_re_time_re_time_request_re_time_re_time_request friendlys.user_id =％s AND friends.friend_id = user.id交叉选择user.id AS user_id，user.username AS user_username，user.email AS user_email，user.password AS user_password，user.first_name AS user_first_name，user.last_name AS user_last_name，user.about_me AS，user_about_me，user.last_seen AS，user_last_seen，user.member_since，AS user_member_since，user.last_message_read_time，AS user_last_message_read_time，user.last_request_read_time，AS user_last_requests_reads_times，\ nFROM用户。 .id）AS anon_1'] [parameters：（1、2）]（此错误的背景位于：http://sqlalche.me/e/f405）