我正在从三个表中获取数据：

$result = $this->db->query("
    SELECT 
        `meetings`.*,
        `follow_up`.id as follow_up_id,
        `follow_up`.comment as follow_up_comment,
        `follow_up`.date as follow_up_date,
        `follow_up`.time as follow_up_time,
        SELECT first_name, last_name, user_mobile, useralt_mobile from users where id = user_id,
        (SELECT address FROM day_location WHERE `meetings`.assigned_to_id = user_id AND `follow_up`.date = date LIMIT 1) AS location_name
        FROM meetings
        LEFT JOIN follow_up ON `meetings`.id = `follow_up`.`meeting_id`
        WHERE follow_up.`date` BETWEEN '{$fromDate_formated}' AND '{$toDate_formated}'
            " . ($user_id > 0 ? " AND `meetings`.assigned_to_id = '{$user_id}'" : '') . "
    ORDER BY `follow_up`.id DESC
");

错误：
发生数据库错误
错误号：1064
sql语法有错误；在第7行的“select first\u name，last\u name，user\u mobile，useralt\u mobile from users where id=”附近，检查与您的mysql服务器版本对应的手册，以获得正确的语法
选择 meetings .*, follow_up .id作为后续id， follow_up .注解作为后续注解， follow_up .日期作为跟进日期， follow_up .time作为follow\u up\u time，选择first\u name、last\u name、user\u mobile、useralt\u mobile from users where id=user\u id，（选择address from day\u location where meetings .assigned_to_id=用户\u id和 follow_up .date=日期限制1）作为地点\u名称从会议左侧加入后续\u meetings .id= follow_up . meeting_id 你在哪里跟进。 date 在“2018-10-01”和“2018-10-31”之间，以及 meetings .assigned \u to \u id='1'订购者 follow_up .id描述
你能帮忙吗？