我在上面的accepted_join_id中发现输入的user_id给出了错误 语法错误，意外的'$ user_id'(T_VARIABLE)
$query=DB::select('select activity_id,accepted_join_id from table_user_create_activity WHERE FIND_IN_SET('$user_id',?) and activity_id=?',[$user_id,$accepted_join_id]);
分析解答
如果你想从php中将$ user_id值注入你的查询，试试这个：
$query=DB::select('select activity_id,accepted_join_id from table_user_create_activity WHERE FIND_IN_SET('.$user_id.',?) and activity_id=?',[$user_id,$accepted_join_id]);
或者如果$ user_id在查询中，它应该像：
$query=DB::select('select activity_id,accepted_join_id from table_user_create_activity WHERE FIND_IN_SET("$user_id",?) and activity_id=?',[$user_id,$accepted_join_id]);