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

如果函数和装饰器都必须接受参数,可以采用下面的方法。

例如,有一个名为decorator1的装饰器,它接受一个参数

@decorator1(5)
def func1(arg1, arg2):
    print (arg1, arg2)

func1(1, 2)

现在,如果decorator1参数必须是动态的,或者在调用函数时传递,

def func1(arg1, arg2):
    print (arg1, arg2)


a = 1
b = 2
seconds = 10

decorator1(seconds)(func1)(a, b)

在上面的代码中

Seconds是decorator1的参数 A b是func1的参数

其他回答

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

如果函数和装饰器都必须接受参数,可以采用下面的方法。

例如,有一个名为decorator1的装饰器,它接受一个参数

@decorator1(5)
def func1(arg1, arg2):
    print (arg1, arg2)

func1(1, 2)

现在,如果decorator1参数必须是动态的,或者在调用函数时传递,

def func1(arg1, arg2):
    print (arg1, arg2)


a = 1
b = 2
seconds = 10

decorator1(seconds)(func1)(a, b)

在上面的代码中

Seconds是decorator1的参数 A b是func1的参数

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。

编写一个带参数和不带参数的装饰器是一个挑战,因为Python在这两种情况下期望完全不同的行为!许多答案都试图解决这个问题,下面是@norok2对答案的改进。具体来说,这种变化消除了locals()的使用。

下面是@norok2给出的相同示例:

import functools

def multiplying(f_py=None, factor=1):
    assert callable(f_py) or f_py is None
    def _decorator(func):
        @functools.wraps(func)
        def wrapper(*args, **kwargs):
            return factor * func(*args, **kwargs)
        return wrapper
    return _decorator(f_py) if callable(f_py) else _decorator


@multiplying
def summing(x): return sum(x)

print(summing(range(10)))
# 45


@multiplying()
def summing(x): return sum(x)

print(summing(range(10)))
# 45


@multiplying(factor=10)
def summing(x): return sum(x)

print(summing(range(10)))
# 450

玩一下这段代码。

问题是用户必须提供键、值对的参数,而不是位置参数,并且第一个参数是保留的。

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