01 - 身份證驗證演算法

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02 - Python實現

 1 #!/usr/bin/python3
 2 """ A simple utility to validate ID for Chinese
 3 
 4     References:
 5     o https://jingyan.baidu.com/article/ff4116259e0a7112e48237b9.html
 6     o https://www.cnblogs.com/xudong-bupt/p/3293838.html
 7 """
 8 
 9 import sys
10 
11 
12 def get_verification_code(id17):
 
13     l_wi = [7, 9, 10, 5, 8, 4, 2, 1, 6, 3, 7, 9, 10, 5, 8, 4, 2]
14     l_vc = ['1', '0', 'X', '9', '8', '7', '6', '5', '4', '3', '2']
15     s = 0
16     for i in range(len(l_wi)):
17         s += l_wi[i] * int(id17[i])
18     m = s % 11
19     return l_vc[m]
20 
21 
22 def is_valid_id(cn_id):
23     cn_id_17 = cn_id[0:-1]
 
24     vc = get_verification_code(cn_id_17)
25     if vc != cn_id[-1]:
26         print("Oops, invalid ID #<<< should be %s%s" % (cn_id_17, vc),
27               file=sys.stderr)
28         return False
29     return True
30 
31 
32 def main(argc, argv):
33     if argc != 2:
34         print
 
("Usage: %s <ID>" % argv[0], file=sys.stderr)
35         print("e.g.", file=sys.stderr)
36         print("       %s 51023519890708911X" % argv[0], file=sys.stderr)
37         return 1
38 
39     cn_id = argv[1]
40     if len(cn_id) != 18:
41         print("-FAIL, invalid ID as its length is not 18", file=sys.stderr)
42         return 1
43 
44     if not is_valid_id(cn_id):
45         print("-FAIL", file=sys.stderr)
46         return 1
47     print("+OK")
48     return 0
49 
50 
51 if __name__ == '__main__':
52     sys.exit(main(len(sys.argv), sys.argv))

執行樣例：

huanli@idorax16:~$ ./validate_id.py 510235198907089112
Oops, invalid ID #<<< should be 51023519890708911X
-FAIL
huanli@idorax16:~$ ./validate_id.py 51023519890708911X
+OK

參考資料：

• 身份證校驗碼計算公式及例項計算
• 身份證號碼驗證演算法