我有一个由装饰器转移变量insurance_mode的问题。我将通过以下装饰器语句来实现:
@execute_complete_reservation(True)
def test_booking_gta_object(self):
self.test_select_gta_object()
但不幸的是,这种说法并不管用。也许也许有更好的办法来解决这个问题。
def execute_complete_reservation(test_case,insurance_mode):
def inner_function(self,*args,**kwargs):
self.test_create_qsf_query()
test_case(self,*args,**kwargs)
self.test_select_room_option()
if insurance_mode:
self.test_accept_insurance_crosseling()
else:
self.test_decline_insurance_crosseling()
self.test_configure_pax_details()
self.test_configure_payer_details
return inner_function
上面的回答很棒。这个例子还演示了@wraps,它从原始函数中获取文档字符串和函数名,并将其应用于新的包装版本:
from functools import wraps
def decorator_func_with_args(arg1, arg2):
def decorator(f):
@wraps(f)
def wrapper(*args, **kwargs):
print("Before orginal function with decorator args:", arg1, arg2)
result = f(*args, **kwargs)
print("Ran after the orginal function")
return result
return wrapper
return decorator
@decorator_func_with_args("foo", "bar")
def hello(name):
"""A function which prints a greeting to the name provided.
"""
print('hello ', name)
return 42
print("Starting script..")
x = hello('Bob')
print("The value of x is:", x)
print("The wrapped functions docstring is:", hello.__doc__)
print("The wrapped functions name is:", hello.__name__)
打印:
Starting script..
Before orginal function with decorator args: foo bar
hello Bob
Ran after the orginal function
The value of x is: 42
The wrapped functions docstring is: A function which prints a greeting to the name provided.
The wrapped functions name is: hello
定义这个decoratorize函数来生成定制的decorator函数:
def decoratorize(FUN, **kw):
def foo(*args, **kws):
return FUN(*args, **kws, **kw)
return foo
可以这样用:
@decoratorize(FUN, arg1 = , arg2 = , ...)
def bar(...):
...
上面的回答很棒。这个例子还演示了@wraps,它从原始函数中获取文档字符串和函数名,并将其应用于新的包装版本:
from functools import wraps
def decorator_func_with_args(arg1, arg2):
def decorator(f):
@wraps(f)
def wrapper(*args, **kwargs):
print("Before orginal function with decorator args:", arg1, arg2)
result = f(*args, **kwargs)
print("Ran after the orginal function")
return result
return wrapper
return decorator
@decorator_func_with_args("foo", "bar")
def hello(name):
"""A function which prints a greeting to the name provided.
"""
print('hello ', name)
return 42
print("Starting script..")
x = hello('Bob')
print("The value of x is:", x)
print("The wrapped functions docstring is:", hello.__doc__)
print("The wrapped functions name is:", hello.__name__)
打印:
Starting script..
Before orginal function with decorator args: foo bar
hello Bob
Ran after the orginal function
The value of x is: 42
The wrapped functions docstring is: A function which prints a greeting to the name provided.
The wrapped functions name is: hello
以下是对t.dubrownik的回答稍加修改的版本。为什么?
作为通用模板,您应该返回原始函数的返回值。
这将改变函数的名称,这可能会影响其他装饰器/代码。
所以使用@functools.wraps():
from functools import wraps
def create_decorator(argument):
def decorator(function):
@wraps(function)
def wrapper(*args, **kwargs):
funny_stuff()
something_with_argument(argument)
retval = function(*args, **kwargs)
more_funny_stuff()
return retval
return wrapper
return decorator
这是curry函数的一个很好的用例。
curry函数本质上是延迟函数的调用,直到提供了所有输入。
这可以用于各种事情,如包装器或函数式编程。在本例中,让我们创建一个接受输入的包装器。
我将使用一个简单的包pamda,其中包含一个用于python的curry函数。这可以用作其他函数的包装器。
安装 Pamda:
pip install pamda
创建一个简单的带有两个输入的装饰函数:
@pamda.curry()
def my_decorator(input, func):
print ("Executing Decorator")
print(f"input:{input}")
return func
使用提供给目标函数的第一个输入应用你的装饰器:
@my_decorator('Hi!')
def foo(input):
print('Executing Foo!')
print(f"input:{input}")
执行你的包装函数:
x=foo('Bye!')
把所有东西放在一起:
from pamda import pamda
@pamda.curry()
def my_decorator(input, func):
print ("Executing Decorator")
print(f"input:{input}")
return func
@my_decorator('Hi!')
def foo(input):
print('Executing Foo!')
print(f"input:{input}")
x=foo('Bye!')
将:
Executing Decorator
input:Hi!
Executing Foo!
input:Bye!