用Python实现查询手机号码归属地实用代码

文章由Byrx.net分享于2019-03-22 12:03:06评论（207）

用Python实现查询手机号码归属地实用代码


用Python实现手机号码归属查询代码

  1. Python 2.6.2 (r262:71600, Aug 4 2009, 09:26:47)
  2. [GCC 3.4.6 20060404 (Red Hat 3.4.6-10)] on linux2
  3. Type "help", "copyright", "credits" or "license" for more information.
  4. >>> from PhoneRegion import getRegion,loadRegion
  5. >>> numsdb = loadRegion()
  6. >>> print getRegion(1234123,numsdb)
  7. 未知,未知
  8. >>> print getRegion(15213492123,numsdb)
  9. 未知,未知
  10. >>> print getRegion(13213492123,numsdb)
  11. 河南濮阳,联通130卡
  12. >>> print getRegion(13513492123,numsdb)
  13. 河北沧州,移动全球通卡
  15. #!/usr/bin/env python
  16. # -*- coding=utf-8 -*-
  18. import re
  19. import sys
  20. import gzip
  21. numberRe = re.compile(^1[0-9]{10}$)
  22. numsDbRe = re.compile((^1[0-9]{6}) (.*) (.*))
  24. def getRegion(number,numsDb):
  25.     if not numberRe.search(number):
  26.         return 未知,未知
  27.     if not numsDb:
  28.         return 未知,未知
  30.     k = number[0:3]
  31.     v = number[3:7]
  32.     if not k in numsDb:
  33.         return 未知,未知
  34.     else:
  35.         region = numsDb[k].get(v,(未知,未知))
  36.         return ,.join(region)
  38. def loadRegion():
  39.     RegionDb = {}
  40.     try:
  41.         for line in gzip.open(numsdb.tar.gz).readlines():
  42.             line = line.strip()
  43.             m = numsDbRe.search(line)
  44.             if m:
  45.                 nums,area,card = m.groups()
  46.                 k = nums[0:3]
  47.                 v = nums[3:7]
  48.                 if k in RegionDb:
  49.                     RegionDb[k][v] = (area,card)
  50.                 else:
  51.                     RegionDb[k]={}
  52.                     RegionDb[k][v] = (area,card)
  53.         return RegionDb
  54.     except Exception,e:
  55.         print e
  56.         return {}
  58. if __name__ == __main__:
  59.     numsDb = loadRegion()
  60.     try:
  61.         number = sys.argv[1]
  62.         print number,getRegion(number,numsDb)
  63.     except:
  64.         pass

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