insert into co_user(mailbox_id,domain_id,realname,engname,oabshow,showorder,eenumber,gender,birthday,homepage,tel_mobile,tel_home,tel_work,tel_work_ext,tel_group,im_qq,im_msn,addr_country,addr_state,addr_city,addr_address,addr_zip,remark,last_session,last_login,openid,unionid,wx_id) values('$mailbox_id','$domain_id','testdel2','NULL','1','0','NULL','male','0000-00-00','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','0','0','0');

2、查詢兩個表mailbox_id的差集。

select * from core_mailbox where mailbox_id not in (select mailbox_id from co_user);

備註（大神方法）：

好吧！這個是可以，但是數據多了的話，想到這個查詢的邏輯有點受不住

於是再改為下面的這樣：

select cu.mailbox_id,cm.mailbox_id  co_user as cu 
　　　　left join core_mailbox as cm 
　　　　　　on cu.mailbox_id = cm.mailbox_id 
　　　　　　　　where cm.mailbox_id  NULL;

利用了left join的，然後進行對比，並且利用where進行篩選。

後面也在網上找了這條：

SELECT mailbox_id FROM `co_user` left join 
(select mailbox_id as i from core_mailbox) as t1
on co_user.mailbox_id= t1.i where t1.i is NULL;

概念上與第二條同理。

LEFT JOIN 關鍵字會從左表 (table_name1) 那裏返回所有的行，即使在右表 (table_name2) 中沒有匹配的行。

msyql命令實例（持續更新）