ATM系统
#coding=utf8
#Version:python 3.6.4
#Tools:Python 2019.9.7
_data_ = '2019/9/7/016 15:27' #时间
_author_ ='gaofeng' #作者
'''
ATM机
需求：
1.登陆
输入账号输入密码
每日只有3次登陆密码错误的机会，超过3次禁止登陆
2.查询余额
3.存款
4.取款
5.转帐
6.退出
'''
infos = [
{'name': 'gaofeng', 'pwd': '123', 'yue': 0},
{'name': 'chenggc', 'pwd': '123', 'yue': 0},
{'name': 'jinyuy', 'pwd': '123', 'yue': 0},
]
# 禁止登陆
ban = False
# 登陆状态
login_state = False
# 账号密码错误次数
login_num = 0
# 菜单 choice
choice = 0
# 输入的用户名
name = ''
# 输入的密码
pwd = ''
# 余额
yue = 0
while ban == False:
# 登陆
while login_state == False:
# 登陆
name = input('请输入账号：')
pwd = input('请输入密码：')
# 查询账号密码是否正确
for info in infos:
if name == info['name'] and pwd == info['pwd']:
print('登陆成功！')
login_state = True
yue = info['yue']
break
# 如果没有账号密码不正确则 运行下面的代码
if login_state == False:
login_num += 1
print('账号密码错误！请重新输入！剩余次数 %s' % (3 - login_num))
if login_num == 3:
print('今日登陆错误次数超限，请明日再来')
ban = True
break
# 打印菜单
if login_state == True and choice == 0:
print('''
1 查询
2 存款
3 转帐
4 取款
5 退出
''')
choice = int(input('请输入操作状态：'))
# 查询
if login_state == True and choice == 1:
print('%s 账户的余额为 %s ¥' % (name, yue))
choice = 0
# 存款
if login_state == True and choice == 2:
money = int(input('请输入你的金额：'))
for info in infos:
if name == info['name']:
info['yue'] += money
yue += money
print('%s 账户存入 %s ¥' % (name, money))
choice = 0
# 转帐:
if login_state == True and choice == 3:
to_state = 0 # 1:成功 2：余额不足
to_num = 0
while to_state == 0:
to_name = input('请输入转入的用户名：')
# 判断 用户是否存在
for info in infos:
if to_name == info['name']:
money = int(input('请输入转帐的金额：'))
# 判断自己账户的金额是否足够
if yue >= money:
info['yue'] += money
yue -= money
print('你给 %s 用户成功转入 %s ¥' % (to_name, money))
to_state = 1
break
else:
print('你的余额不足，请充值后再转帐')
to_state = 2
if to_state == 1:
break
elif to_state == 2:
break
else:
to_num += 1
print('你输入的用户不存在,请重新输入：')
choice = 0
# 取款
if login_state == True and choice == 4:
money = int(input('请输入你的取款金额：'))
# 判断输入的金额是否小于等于余额
if money <= yue:
print('请再出钞口取钞票吧！')
yue -= money
choice = 0
else:
print('你输入的金额大于你的余额，无法取款！')
# 退出
if login_state == True and choice == 5:
break
购物车系统
#coding=utf8
#Version:python 3.6.4
#Tools:Python 2019.9.7
_data_ = '2019/9/7/016 15:27' #时间
_author_ ='gaofeng' #作者
salary = int(input("请输入你的工资:"))
shoppingmart = []
items = (["1","huawei","￥",6799],["2","earphone","￥",3000],["3","notebook","￥",8000])
msg_items = """
--------items-------
1.huawei $ 6799
2.earphone $ 3000
3.notebook $ 8000
-------------------
"""
print(msg_items)
while True:
shopindex = int(input("选择商品:"))
if salary > items[shopindex-1][3]:
shoppingmart.append(items[shopindex-1])
salary -=int(items[shopindex-1][3])
print("你已经买了 {name}!".format(name=items[shopindex-1][1]))
print("你的余额是:$",salary)
decision =input("你现在想离开吗?")
print(msg_items)
else:
print("你的余额不够!请试试别的.")
recharge_ans = input("你想充值码?")
if recharge_ans =="y":
recharge = int(input("请输入金额:"))
print("请等一会...")
salary +=recharge
print("你充值成功了!")
print("余额是:",salary,"现在!")
decision =input("你现在想离开吗?")
print(msg_items)
if decision == "q":
break
else:
continue
print("你已经买了:",shoppingmart)
print("你的余额:$,salary")
print("欢迎你下次光临!")