很多人用不同的东西来解决易语言难题，咱也凑个热闹。如果用存储过程的话，和一般的编程语言也就雷同了，这里就用SQL的思想，在MySQL中实现一次。 MySQL drop table if exists n1;create temporary table n1(num int (1));insert into n1 values(1), (2), (3)
很多人用不同的东西来解决易语言难题，咱也凑个热闹。如果用存储过程的话，和一般的编程语言也就雷同了，这里就用SQL的思想，在MySQL中实现一次。
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("[", n3.num, "]");
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("[", n2.num, "]")
-> and n1.num not rlike concat("[", n3.num, "]")
-> and n2.num not rlike concat("[", n3.num, "]");
+------+------+------+
| num | num | num |
+------+------+------+
| 192 | 384 | 576 |
| 219 | 438 | 657 |
| 273 | 546 | 819 |
| 327 | 654 | 981 |
+------+------+------+
4 rows in set (0.03 sec)
