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。