函数回顾05

login_dic = {"username": None, "flag": False} msg = """请选择app: QQ 微信 抖音 邮箱 """

chose = input(msg).upper() def auth(argv): def wrapper(f): def inner(*args,**kwargs): if login_dic["flag"]: f(*args,**kwargs) else: if argv == "QQ": print("欢迎登陆QQ") user = input("请输入账号:") password = input("请输入密码:") if user == "alex" and password == "alex123": login_dic["flag"] = True login_dic["username"] = user f(*args,**kwargs) else: print("账号或者密码错误") elif argv == "微信": print("欢迎登陆微信") user = input("请输入账号:") password = input("请输入密码:") if user == "alex123" and password == "alex123456": login_dic["flag"] = True login_dic["username"] = user f(*args, **kwargs) else: print("账号或者密码错误") elif argv == "抖音": print("欢迎登陆抖音") user = input("请输入账号:") password = input("请输入密码:") if user == "alex456" and password == "alex456123": login_dic["flag"] = True login_dic["username"] = user f(*args, **kwargs) else: print("账号或者密码错误") else: print("欢迎登陆邮箱") user = input("请输入账号:") password = input("请输入密码:") if user == "alex@qq.com" and password == "alex123": login_dic["flag"] = True login_dic["username"] = user f(*args, **kwargs) else: print("账号或者密码错误") return inner return wrapper

@auth(chose) def foo(): print("这是一个被装饰的函数") foo()

@auth(chose) 相等于以下两行代码的解构

wrapper = auth(chose)

foo = wrapper(foo)

2.多个装饰器装饰一个函数 def wrapper1(func): def inner1(*args,**kwargs): print(1) func(*args,**kwargs) print(11) return inner1

def wrapper2(func): def inner2(*args,**kwargs): print(2) func(*args,**kwargs) print(22) return inner2

def wrapper3(func): def inner3(*args,**kwargs): print(3) func(*args,**kwargs) print(33) return inner3

@wrapper3 @wrapper2 @wrapper1 def foo(): print("这是一个被装饰的函数")

foo()

3

2 1 这是一个被装饰的函数 11 22 33

posted on 2019-07-24 21:20  七橼77 阅读( ...) 评论( ...) 编辑 收藏

转载于:https://www.cnblogs.com/-777/p/11240930.html