MySQL (非存储过程)解决易语言难题
发布时间:2021-01-14 17:04:50 所属栏目:MySql教程 来源:网络整理
导读:今天PHP站长网 52php.cn把收集自互联网的代码分享给大家,仅供参考。 drop table if exists n1;create temporary table n1(num int (1));insert into n1 values(1),(2),(3),(4),(5),(6),(7),(8),(9);drop table if exists
以下代码由PHP站长网 52php.cn收集自互联网 现在PHP站长网小编把它分享给大家,仅供参考 drop table if exists n1; create temporary table n1(num int (1)); insert into n1 values(1),(2),(3),(4),(5),(6),(7),(8),(9); drop table if exists n2; create temporary table n2 select * from n1; drop table if exists n3; create temporary table n3 select * from n1; drop table if exists nums1; create temporary table nums1 select n1.num * 100 + n2.num * 10 + n3.num as num from n1 left join n2 on n1.num <> n2.num left join n3 on n1.num <> n3.num and n2.num <> n3.num; drop table if exists nums2; create temporary table nums2 select * from nums1; drop table if exists nums3; create temporary table nums3 select * from nums1; select * from nums1 as n1 left join nums2 as n2 on n1.num <> n2.num left join nums3 as n3 on n1.num <> n3.num and n2.num <> n3.num where n1.num * 2 = n2.num and n1.num * 3 = n3.num and n1.num not rlike concat("[",n2.num,"]") and n1.num not rlike concat("[",n3.num,"]") and n2.num not rlike concat("[","]"); drop table if exists n1; drop table if exists n2; drop table if exists n3; drop table if exists nums1; drop table if exists nums2; drop table if exists nums3; 结果: mysql> select * -> from nums1 as n1 left join nums2 as n2 -> on n1.num <> n2.num -> left join nums3 as n3 -> on n1.num <> n3.num and n2.num <> n3.num -> where n1.num * 2 = n2.num and n1.num * 3 = n3.num -> and n1.num not rlike concat("[","]") -> and n1.num not rlike concat("[","]") -> and n2.num not rlike concat("[","]"); +------+------+------+ | num | num | num | +------+------+------+ | 192 | 384 | 576 | | 219 | 438 | 657 | | 273 | 546 | 819 | | 327 | 654 | 981 | +------+------+------+ 4 rows in set (0.03 sec) 以上内容由PHP站长网【52php.cn】收集整理供大家参考研究 如果以上内容对您有帮助,欢迎收藏、点赞、推荐、分享。 (编辑:好传媒网) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |