Python实现CET查分的方法

Python CET自动查询方法需要用到的python方法模块有：sys、urllib2

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：

#!/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。