MySQL （非存储过程）解决易语言难题

永久禁言

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。

1. drop table if exists n1;
   
2. create temporary table n1(num int (1));
   
3. insert into n1 values(1), (2), (3), (4), (5), (6), (7), (8), (9);
   
4. 
   
5. drop table if exists n2;
   
6. create temporary table n2
   
7. select * from n1;
   
8. 
   
9. drop table if exists n3;
   
10. create temporary table n3
    
11. select * from n1;
    
12. 
    
13. drop table if exists nums1;
    
14. create temporary table nums1
    
15. select n1.num * 100 + n2.num * 10 + n3.num as num
    
16. from n1 left join n2
    
17. on n1.num <> n2.num
    
18. left join n3
    
19. on n1.num <> n3.num and n2.num <> n3.num;
    
20. 
    
21. drop table if exists nums2;
    
22. create temporary table nums2
    
23. select * from nums1;
    
24. 
    
25. drop table if exists nums3;
    
26. create temporary table nums3
    
27. select * from nums1;
    
28. 
    
29. select *
    
30. from nums1 as n1 left join nums2 as n2
    
31. on n1.num <> n2.num
    
32. left join nums3 as n3
    
33. on n1.num <> n3.num and n2.num <> n3.num
    
34. where n1.num * 2 = n2.num and n1.num * 3 = n3.num
    
35. and n1.num not rlike concat("[", n2.num, "]")
    
36. and n1.num not rlike concat("[", n3.num, "]")
    
37. and n2.num not rlike concat("[", n3.num, "]");
    
38. 
    
39. drop table if exists n1;
    
40. drop table if exists n2;
    
41. drop table if exists n3;
    
42. drop table if exists nums1;
    
43. drop table if exists nums2;
    
44. drop table if exists nums3;
    
45. 
    
46. 
    
47. 结果：
    
48. mysql> select *
    
49.     -> from nums1 as n1 left join nums2 as n2
    
50.     -> on n1.num <> n2.num
    
51.     -> left join nums3 as n3
    
52.     -> on n1.num <> n3.num and n2.num <> n3.num
    
53.     -> where n1.num * 2 = n2.num and n1.num * 3 = n3.num
    
54.     -> and n1.num not rlike concat("[", n2.num, "]")
    
55.     -> and n1.num not rlike concat("[", n3.num, "]")
    
56.     -> and n2.num not rlike concat("[", n3.num, "]");
    
57. +------+------+------+
    
58. | num  | num  | num  |
    
59. +------+------+------+
    
60. |  192 |  384 |  576 |
    
61. |  219 |  438 |  657 |
    
62. |  273 |  546 |  819 |
    
63. |  327 |  654 |  981 |
    
64. +------+------+------+
    
65. 4 rows in set (0.03 sec)
    

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