編寫登陸接口
基礎需求：
1. 讓用戶輸入用戶名密碼
2. 認證成功後顯示歡迎信息
3. 輸錯三次後退出程序
升級需求：
4. 可以支持多個用戶登錄 (提示，通過列表存多個賬戶信息)
5. 用戶3次認證失敗後，退出程序，再次啟動程序嘗試登錄時，還是鎖定狀態（提示:需把用戶鎖定的狀態存到文件裏）

#!/usr/bin/env python
# -*- coding:utf-8 -*-  
# by wk

‘‘‘
說明：
如果登錄用戶在用戶列表裏，每個用戶只有3次登錄機會，失敗後鎖定賬戶，下次啟動賬戶提示賬戶被鎖定
如果登錄用戶不在用戶列表裏，提示用戶不存在,並且嘗試3次登錄，如果失敗退出程序
 
‘‘‘

import sys

def checklock(username):
    with open("loginnum.txt", ‘r‘) as lock_t:
        for line in lock_t.readlines():
            if len(line) == 0:
                continue
            if username == line.strip():
                print("login fail too much! User locked")
                # sys.exit()
 

                return ‘lock‘


def userlock(username):
    f = open("loginnum.txt", ‘a‘)
    f.write(username + ‘\n‘)
    f.close()


def userlogin(userinfo):
    count = 3                    #計數器
    flag = ‘success‘             #標記登錄狀態
    userid = 0                   #標記用戶編號
    while True:
        
 
# if userinfo[userid][‘login_fail‘] != 0:
        username = input(‘please enter your name: ‘)
        password = input(‘please enter your password: ‘)
        if checklock(username) == ‘lock‘:
            continue
        for i,user in enumerate(userinfo):
            if user[‘username‘] == username and user[‘password‘] == password:
                flag = ‘success‘
                break
            elif user[‘username‘] == username and user[‘password‘] != password:
                flag = ‘fail‘
                userid = int(i)
                break
            else:
               flag = ‘nouser‘
        if flag == ‘success‘:
            print(‘login successful, welcome!‘)
            break
        elif flag == ‘fail‘:
            # count -= 1
            print(‘who ‘, userinfo[userid])
            failnum = int(userinfo[userid][‘login_fail‘])             #統計登錄失敗次數
            failnum -= 1
            if userinfo[userid][‘login_fail‘] == 1:
                print(‘your account lock‘)
                userlock(userinfo[userid][‘username‘])
                sys.exit()
            else:
                userinfo[userid][‘login_fail‘] = failnum
                print(‘login fail, you have %s choice‘ % (failnum))
        else:
            count -= 1
            if count == 0:
                print(‘you login too much time‘)
                sys.exit()
            print(‘Sorry no user, you have %s choice‘ % (count))
        # else:
        #     break

if __name__ == ‘__main__‘:
    userinfo = [
        {‘username‘:‘eric‘, ‘password‘:‘123456‘, ‘login_fail‘:‘3‘},
        {‘username‘:‘tom‘, ‘password‘:‘123456‘, ‘login_fail‘:‘3‘},
        {‘username‘:‘jerry‘, ‘password‘:‘123456‘, ‘login_fail‘:‘3‘},
    ]
    # password = [‘123456‘,‘123456‘,‘123456‘]
    userlogin(userinfo)

猜年齡遊戲升級版
要求：
1. 允許用戶最多嘗試3次
2. 每嘗試3次後，如果還沒猜對，就問用戶是否還想繼續玩，如果回答Y或y, 就繼續讓其猜3次，以此往復，如果回答N或n，就退出程序
3. 如何猜對了，就直接退出

#!/usr/bin/env python
# -*- coding:utf-8 -*-  
# by wk

‘‘‘
說明：
1. 允許用戶最多嘗試3次
2. 每嘗試3次後，如果還沒猜對，就問用戶是否還想繼續玩，如果回答Y或y, 就繼續讓其猜3次，以此往復，如果回答N或n，就退出程序
3. 如何猜對了，就直接退出

‘‘‘

def guessage(my_age):
    flag = ‘‘
    while True:
        count = 3
        for i in range(3):                  #循環3次
            try:
                age = int(input(‘please enter age: ‘))
                if age > my_age:
                    count -= 1              #統計輸錯次數
                    print(‘you guess too big, you have %s chance‘ %count)
                    if i == 2:              #猜錯3次後詢問還繼續猜不
                        # print(‘you have no chioce‘)
                        flag = input(‘Do you want try again Y/N ? ‘)
                elif age < my_age:
                    count -= 1              #統計輸錯次數
                    print(‘you guess too smaill, you have %s chance‘ %count)
                    if i == 2:              #猜錯3次後詢問還繼續猜不
                        # print(‘you have no chioce‘)
                        flag = input(‘Do you want try again Y/N ? ‘)
                else:
                    print(‘you guess it !‘)      #猜對了退出
                    flag = ‘N‘
                    break;
            except:                         #如果輸入的不是數字，提示請輸入數字
                count -= 1
                print(‘please enter number, you have %s chance‘ %count)
        if flag == ‘Y‘ or flag == ‘y‘:
            continue
        elif flag == ‘N‘ or flag == ‘n‘:
            break
        else:
            print(‘Wrong choice !!!‘)       #如果輸入的不是Y,y,N,n,直接退出
            break

if __name__ == ‘__main__‘:
    my_age = 32
    guessage(my_age)

登錄接口，猜年齡