MySQL （非存储过程）解决易语言难题

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：清远站长网）

【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!