Php 使用join或alias获取具有相同报告电子邮件地址的用户的详细信息

我有两张桌子

tbl_用户：

id fisrtname lastname
5   John      Doe
6   Peter     Parker
7   Will      Smith

tbl_经验：

exp_id usr_id user_reporting_to
1       5     dev@abc.com 
2       6     admin@abc.com
3       7     dev@abc.com

我想获取那些与where条件usr_id=5具有相同报告电子邮件地址的记录，因为当我登录到我的页面时，它会创建用户id会话，例如，现在使用join或alias的where条件为5

SELECT texperience.tbl_experience_report_to_email AS tbl_experience_report_to_email, 
    tuser.tbl_user_fname AS tbl_user_fname, 
    texperience.tbl_experience_designation AS tbl_experience_designation 
FROM tbl_experience AS texperience,
    tbl_user AS tuser 
WHERE tuser.tbl_user_fname = tuser.tbl_experience_id 
    AND texperience.tbl_experience_report_to_email = texperience.tbl_experience_id
    AND texperience.tbl_experience_user_id = 1

基本上有一个子查询，将所有电子邮件地址分组，并且联接表将返回那些在tbl_体验中多次出现电子邮件的用户

    SELECT tbl_user.*, tbl_experience.user_reporting_to FROM tbl_user 
    LEFT JOIN tbl_experience ON (tbl_user.id=tbl_experience.usr_id) 
    LEFT JOIN (SELECT count(*) as total_user, user_reporting_to FROM tbl_experience GROUP BY user_reporting_to) as email_group ON 
(email_group.user_reporting_to = tbl_experience.user_reporting_to) WHERE email_group.total_user > 1