Python基础-02

                                                                              Python基础-02

本节内容:

1、         三级菜单

2、         字符编码

一、    三级菜单

#要求 1。可以一层一层进入到所有层
#    2。可以在每一层返回上一层
#    3.可以在任意层退出主菜单
#    4.写重复代码写成不重复

无脑版(单纯按照思路一步一步写)

back_flag = False
exit_flag = False

while not back_flag and not exit_flag:
    for key in manu:#打印第一层 (循环第一层)
        print(key)
    choice = input(">>:").strip()
    if choice == "q":
        exit_flag = True
    if choice in manu:#打印第二层
        while not back_flag and not exit_flag:#让程序停留在第二层
            for key2 in manu[choice]:
                print(key2)
            choice2 = input(">>:").strip()
            if choice2 == "b":
                back_flag = True
            if choice2 == "q":
                back_flag = True
            if choice2 in manu[choice]:
                while not back_flag and not exit_flag:#False
                    for key3 in manu[choice][choice2]:
                        print(key3)
                    choice3 = input("3>>:").strip()
                    if choice3 == "b":
                        back_flag = True
                    if choice3 == "q":
                        exit_flag = True
                    if choice3 in manu[choice][choice2]:
                        while not back_flag and not exit_flag:
                            for key4 in manu[choice][choice2][choice3]:
                                print(key4)
                            choice4 = input("4>>:").strip()
                            print("last level")
                            if choice4 == "b":
                                back_flag=True
                            if choice4 == "q":
                                exit_flag=True
                        else:
                            back_flag = False
                else:
                    back_flag = False
        else:
            back_flag = False

减少重复代码实现三级菜单

menu = {
    '北京':{
        '朝阳':{
            '国贸':{
                'CICC':{},
                'HP': {},
                '渣打银行': {},
                'CCTV': {},

            },
            '望京':{
                '优衣库':{},
                'apple': {},
            },
        },
        '昌平':{
            '沙河':{
                '老男孩':{},
                '阿泰包子':{},
            },
            '天通苑': {
                '链家':{},
                '我爱我家':{},
            },
            '回龙观':{},
        },
        '海淀':{
            '五道口':{
                '谷歌':{},
                '网易':{},
                'Sohu':{},
                'Sogo':{},
                '快手': {},

            },
            '中关村':{
                'Youku':{},
                "Iqiyi":{},
                "汽车之家": {},
                "新东方": {},
                "QQ": {},

            },
        },
    },
    '上海':{
        "浦东":{
            "陆家嘴":{
                "CICC":{},
                "高盛": {},
                "摩根": {},
            },
            "外滩":{}
        },
        "闵行":{},
        "静安":{},
    },
    '山东':{
        "济南":{},
        "德州":{
            "乐陵":{
                "丁务镇":{},
                "城区":{}
            },
            "平原":{}
        },
        "青岛": {},

    },
}

current_layer = menu #实现动态循环
# parent_layer = menu

parent_layers = []#保持所以父级 ,最后一个元素永远都 是父亲级

while True:
    for key in current_layer:
        print(key)
    choice = input(">>>:").strip()
    if len(choice) == 0:continue
    if choice in current_layer:#改之前相当于父亲
        parent_layers.append(current_layer)#在进入下一层之前,把当前层追加到父级
        #下一次loop  当用户选择b的选择,就可以直接取列表的最后一个值出来就ok了
        current_layer = current_layer[choice]#改成了子层
    elif choice == "b":
        # current_layer = parent_layer
        if parent_layers:#判断是否为空
            current_layer = parent_layers.pop()#取出列表最后一个值,因为他就是当前的父值
    else:
        print("无此项")

二、         字符编码

二进制
---》ASCII:智能存英文和拉丁字符,一个字符占一个字节:8位
in python2
# 默认是 ASCII
#in py3
#默认是 unicode
#encode 在编码的同时,会把数据转成bytes类型
#decode 在解码的同时会把bytes类型转成字符串
#b = byte = 字节类型 =【0-255】

例程

s = '特斯拉'
s_to_gbk = s.encode('gbk')
print(s)
print(s_to_gbk)


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转载自blog.csdn.net/qq_27262727/article/details/78994625
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