Python 实现快递查询

实现效果:

源代码：

import urllib.request
import json
import msvcrt

kd_dict = {1:'shentong',2:'youzhengguonei',3:'yuantong',4:'shunfeng',5:'yunda',6:'zhongtong',7:"tiantian",8:"debang"}

def Check():
    while True:
        print("仅支持以下快递公司查询:")
        print("1.申通    ")
        print("2.EMS邮政    ")
        print("3.圆通    ")
        print("4.顺风    ")
        print("5.韵达    ")
        print("6.中通    ")
        print("7.天天    ")
        print("8.德邦    ")
        print("0.退出
")
        choose = int(input("请选择您的快递公司:"))
        while choose not in range(0,9):
            choose = int(input("抱歉暂不支持此公司请重新选择:"))
        if choose == 0:
            print("感谢使用!
")
            break
        kd_num = input("请输入快递单号:")
        url = "http://www.kuaidi100.com/query?type=%s&postid=%s" % (kd_dict[choose], kd_num)
        response = urllib.request.urlopen(url)
        html = response.read().decode('utf-8')
        target = json.loads(html)
        #print(target)
        status = target['status']
        if status == '200':
            data = target['data']
            #print(data)
            data_len = len(data)
            #print(data_len)
            #print("
")
            for i in range(data_len):
                print("
时间: " + data[i]['time'])
                print("状态: " + data[i]['context'] + "")
            print("
感谢使用!
")
            break
        else:
            print("输入有误请重新输入!
")
    #print("按任意键结束......")


if __name__ == '__main__':
    while True:
        Check()
        out = input("按任意数字退出(其他键继续).........")
        if out >= '0' and out <= '9':
            break
        else:
            print("
")
            continue
            

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无聊............