python实现cet查分的方法

本文实例讲述了python实现cet查分的方法。分享给大家供大家参考。具体实现方法如下:

代码如下:

#!/usr/bin/python
# -*- coding: utf-8 -*-
import sys, urllib2
def cetquery(band, exam_id):
“””cetquery version 0.2 2009.2.28
an exercise program by pt, gz university
author blog: http://apt-blog.co.cc , welcome to drop by.
“””
#查询连接
cet = “http://cet.99sushe.com/cetscore_99sushe0902.html?t=” + band + “&id=” + exam_id
print “connecting…”
#构造http头
header = {‘referer’:’http://cet.99sushe.com/’}
#第二个参数出现则使用post方式提交
req = urllib2.request(cet, ”, header)
try:
data = urllib2.urlopen(req).read()
except baseexception, e:
print “error retrieving data:”, e
return -1
if not len(result):
print “error occured. maybe record not existed.”
return -1
#解码字符串
result = data.decode(“gb2312”).encode(“utf8”)
res_tu = tuple(result.split(‘,’))
score_tu = (“听力”, “阅读”, “综合”, “写作”, “总分”, “学校”, “姓名”)
print “n***** cet %s 成绩清单 *****” % (band)
print “-准考证号: %s” % (exam_id)
for i in range(7):
print “-%s: %s” % (score_tu, res_tu)
print “**************************n”
print “准考证号前一位同学: %sn后两位同学分别是: %s、%s” % (res_tu[-3], res_tu[-2], res_tu[-1])
return 0
if __name__ == “__main__”:
if (len(sys.argv) != 3) or
(sys.argv[1] != ‘4’ and sys.argv[1] != ‘6’) or
(len(sys.argv[2]) != 15):
print “error: 程序参数错误,考试类型(4、6),准考证号长度(15位)”
print “nexample:nncetquery.py 4 123456789012345nn”
print cetquery.__doc__
sys.exit(1)
statue = cetquery(sys.argv[1], sys.argv[2])
sys.exit(statue)

希望本文所述对大家的python程序设计有所帮助。

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