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

当前回答

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

其他回答

编辑:为了深入了解装饰师的心理模型,请看看这个很棒的Pycon Talk。这30分钟很值得。

考虑带参数的装饰器的一种方式是

@decorator
def foo(*args, **kwargs):
    pass

翻译为

foo = decorator(foo)

如果decorator有参数,

@decorator_with_args(arg)
def foo(*args, **kwargs):
    pass

翻译为

foo = decorator_with_args(arg)(foo)

Decorator_with_args是一个函数,它接受自定义参数并返回实际的装饰器(将应用于被装饰的函数)。

我使用了一个简单的技巧与部分,使我的装饰容易

from functools import partial

def _pseudo_decor(fun, argument):
    def ret_fun(*args, **kwargs):
        #do stuff here, for eg.
        print ("decorator arg is %s" % str(argument))
        return fun(*args, **kwargs)
    return ret_fun

real_decorator = partial(_pseudo_decor, argument=arg)

@real_decorator
def foo(*args, **kwargs):
    pass

更新:

上面,foo变成了real_decorator(foo)

修饰函数的一个效果是,foo的名字在修饰器声明中被重写。Foo被real_decorator返回的任何东西“覆盖”。在本例中,是一个新的函数对象。

foo的所有元数据都会被重写,尤其是文档字符串和函数名。

>>> print(foo)
<function _pseudo_decor.<locals>.ret_fun at 0x10666a2f0>

functools。Wraps为我们提供了一个方便的方法,将文档字符串和名称“提升”到返回的函数中。

from functools import partial, wraps

def _pseudo_decor(fun, argument):
    # magic sauce to lift the name and doc of the function
    @wraps(fun)
    def ret_fun(*args, **kwargs):
        # pre function execution stuff here, for eg.
        print("decorator argument is %s" % str(argument))
        returned_value =  fun(*args, **kwargs)
        # post execution stuff here, for eg.
        print("returned value is %s" % returned_value)
        return returned_value

    return ret_fun

real_decorator1 = partial(_pseudo_decor, argument="some_arg")
real_decorator2 = partial(_pseudo_decor, argument="some_other_arg")

@real_decorator1
def bar(*args, **kwargs):
    pass

>>> print(bar)
<function __main__.bar(*args, **kwargs)>

>>> bar(1,2,3, k="v", x="z")
decorator argument is some_arg
returned value is None
def decorator(argument):
    def real_decorator(function):
        def wrapper(*args):
            for arg in args:
                assert type(arg)==int,f'{arg} is not an interger'
            result = function(*args)
            result = result*argument
            return result
        return wrapper
    return real_decorator

装饰器的使用

@decorator(2)
def adder(*args):
    sum=0
    for i in args:
        sum+=i
    return sum

然后

adder(2,3)

生产

10

but

adder('hi',3)

生产

---------------------------------------------------------------------------
AssertionError                            Traceback (most recent call last)
<ipython-input-143-242a8feb1cc4> in <module>
----> 1 adder('hi',3)

<ipython-input-140-d3420c248ebd> in wrapper(*args)
      3         def wrapper(*args):
      4             for arg in args:
----> 5                 assert type(arg)==int,f'{arg} is not an interger'
      6             result = function(*args)
      7             result = result*argument

AssertionError: hi is not an interger

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

在我的实例中,我决定通过一行lambda来解决这个问题,以创建一个新的decorator函数:

def finished_message(function, message="Finished!"):

    def wrapper(*args, **kwargs):
        output = function(*args,**kwargs)
        print(message)
        return output

    return wrapper

@finished_message
def func():
    pass

my_finished_message = lambda f: finished_message(f, "All Done!")

@my_finished_message
def my_func():
    pass

if __name__ == '__main__':
    func()
    my_func()

执行时,输出:

Finished!
All Done!

也许不像其他解决方案那样可扩展,但对我来说是可行的。

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(...):
        ...