分享程序员开发的那些事...
更新时间：2022-09-23 11:23:09
1、插入数据
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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');
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2、查询两个表mailbox_id的差集。
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select * from core_mailbox where mailbox_id not in (select mailbox_id from co_user);
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备注（大神方法）：
好吧！这个是可以，但是数据多了的话，想到这个查询的逻辑有点受不住
于是再改为下面的这样：
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，然后进行对比，并且利用where进行筛选。
后面也在网上找了这条：
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;
概念上与第二条同理。
LEFT JOIN 关键字会从左表 (table_name1) 那里返回所有的行，即使在右表 (table_name2) 中没有匹配的行。
本文转自 sailikung 51CTO博客，原文链接：http://blog.51cto.com/net881004/2054822，如需转载请自行联系原作者