标签:[] opp == ftime append 循环 %x time import
要求:
1.用户输入金额后,就可以选择商品添加进购物车
2.金额不足扣款时,提示用户余额不足
3.当用户随时可以退出程序
4.退出程序后自动保存金额和所购商品
# -*- coding:utf-8 -*- import time #分别打开商品列表的储存文件和余额列表的储存文件 with open (‘ware_list‘,‘r‘,encoding = ‘utf-8‘)as ware: ware = eval(ware.read()) with open (‘balance‘,‘r+‘,encoding = ‘utf-8‘) as balance: balance = eval(balance.read()) shopping_cart = [] warelist = [] #根据商品数量打印商品列表 def ware_list(): print(‘---------商品列表-----------‘) count = 0 for i in ware: count += 1 print(count,‘.‘,i,‘ ‘,ware[i]) if i in warelist: continue else: warelist.append(i) def timmer(): time.format =‘%Y-%m-%d %X‘ time.current = time.strft(time.format) print(time.current) #判断是否有余额文件如果有则将余额值赋予工资变量 if balance[‘balance‘] > 0: print((‘well come back! ur balance :{0} ur bought list :{1}‘).format(balance[‘balance‘],balance[‘shopping cart‘])) shopping_cart.extend(balance[‘shopping cart‘]) salary = balance[‘balance‘] else: salary = input(‘请输入您的工资:‘) while True: salary = int(salary) ware_list() choice = input(‘你想购买什么商品?\n (输入Q退出)‘) #如果输入为q则可以退出 if choice.upper() == ‘Q‘: request = input(‘请确认是否退出?Y/N‘) # if request.upper() == ‘Y‘: break elif request.upper()== ‘N‘: continue else: print(‘非法输入!!‘) continue elif len(choice)<= len(warelist): if salary - int(ware[warelist[int(choice)-1]]) > 0: shopping_cart.append(warelist[int(choice)-1]) salary -= int(ware[warelist[int(choice)-1]]) print((‘您已经购买:{0} 余额为:{1}‘).format(shopping_cart,salary)) time.format =‘%Y-%m-%d %X‘ time.current = time.strftime(time.format) print(time.current) else: print(‘您的余额不足购买该商品 请重新选择‘) else: print(‘请正确输入商品序号!‘) continue #保存余额信息 print((‘您一共购买了{0}件商品,您的余额剩余:{1} 购买了:{2}‘).format(len(shopping_cart),salary,shopping_cart)) with open("balance",‘w‘,encoding=‘utf-8‘)as new: new.write("{‘balance‘:%d,‘shopping cart‘:%s}"%(salary,shopping_cart))
缺陷:无法针对回车等非法输入进行限制
思路:因为要求随时可以退出程序 所以减少循环层数是减少工作量提高代码运行效率的最好方式 所以我就只用了一个主循环
标签:[] opp == ftime append 循环 %x time import
原文地址:http://www.cnblogs.com/Lowsheep/p/6834526.html