有参装饰器多个装饰器装饰一个

小咪咪 2023-10-10 10:16 37阅读 0赞

# 1.有参装饰器 #

### 基本版 ###

def auth(argv): def wrapper(f): def inner(*args,**kwargs): f(*args,**kwargs) return inner return wrapper

### 示例: ###

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

转载于:https://www.cnblogs.com/beichen123/p/11253054.html