php - 如何将一个查询结果减去mysql中的另一个查询

Question

这是我的第一个问题

SELECT DISTINCT(u.nickname) FROM user u 
                           where   u.id  IN(SELECT `submission_user`.`user_id` from `submission_user`) AND
                           u.member_since   >= '2015-07-01'

这是我的第二个查询

SELECT DISTINCT(u.nickname) FROM user u LEFT JOIN submission_user su ON su.user_id = u.id  
                            LEFT JOIN submission s ON s.id = su.submission_id
                             WHERE  s.date_time BETWEEN '2017-10-31' and '2018-07-31'

这是我的第三个查询

SELECT DISTINCT(u.nickname) FROM user u LEFT JOIN submission_user su ON su.user_id = u.id  
                            LEFT JOIN track_user tu ON tu.user_id = u.id
                            LEFT JOIN track ON track.id = tu.track_id                       
                            where track.uploaded_timestamp BETWEEN '2017-10-31' and '2018-07-31'

然后，我将合并第二个和第三个查询结果

 $ids_reactivated = array_unique(array_merge($track_user, $submit_user));

所以我的问题是，如果我想减去查询一个结果以合并结果意味着与(查询 2 和 3)即在我的情况下:$ids_reactivated

谁知道怎么做...我已经尝试了很多方法并通过了一天...

希望大家帮帮我谢谢

Answer 1

谢谢你的提示，但我得到了这样的答案......

   SELECT DISTINCT(u.nickname) FROM user u 
                               where   u.id  IN(SELECT `submission_user`.`user_id` from `submission_user`) AND
                               u.member_since   >= '2015-07-01' 
    **and u.nickname not in ($query2)**

UNION 
SELECT DISTINCT(u.nickname) FROM user u 
                               where   u.id  IN(SELECT `submission_user`.`user_id` from `submission_user`) AND
                               u.member_since   >= '2015-07-01' 
    **and u.nickname not in ($query3)**