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

例如,我在下面创建了multiply(),它可以接受一个参数或不接受参数,也可以不接受装饰器的括号,我在下面创建了sum():

from numbers import Number

def multiply(num=1):
    def _multiply(func):
        def core(*args, **kwargs):
            result = func(*args, **kwargs)
            if isinstance(num, Number):
                return result * num
            else:
                return result
        return core
    if callable(num):
        return _multiply(num)
    else:
        return _multiply

def sum(num1, num2):
    return num1 + num2

现在,我把@multiply(5)放在sum()上,然后调用sum(4,6),如下所示:

# (4 + 6) x 5 = 50

@multiply(5) # Here
def sum(num1, num2):
    return num1 + num2

result = sum(4, 6)
print(result)

那么,我可以得到如下结果:

50

接下来,我把@multiply()放在sum()上,然后调用sum(4,6),如下所示:

# (4 + 6) x 1 = 10

@multiply() # Here
def sum(num1, num2):
    return num1 + num2
    
result = sum(4, 6)
print(result)

或者,我把@multiply放在sum()上,然后调用sum(4,6),如下所示:

# 4 + 6 = 10

@multiply # Here
def sum(num1, num2):
    return num1 + num2
    
result = sum(4, 6)
print(result)

那么,我可以得到如下结果:

10

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

带参数的装饰器的语法有点不同——带参数的装饰器应该返回一个函数,该函数将接受一个函数并返回另一个函数。它应该返回一个普通的装饰器。有点困惑,对吧?我的意思是:

def decorator_factory(argument):
    def decorator(function):
        def wrapper(*args, **kwargs):
            funny_stuff()
            something_with_argument(argument)
            result = function(*args, **kwargs)
            more_funny_stuff()
            return result
        return wrapper
    return decorator

在这里你可以读到更多关于这个主题的内容——也可以使用可调用对象来实现这个功能,这里也有解释。

例如,我在下面创建了multiply(),它可以接受一个参数或不接受参数,也可以不接受装饰器的括号,我在下面创建了sum():

from numbers import Number

def multiply(num=1):
    def _multiply(func):
        def core(*args, **kwargs):
            result = func(*args, **kwargs)
            if isinstance(num, Number):
                return result * num
            else:
                return result
        return core
    if callable(num):
        return _multiply(num)
    else:
        return _multiply

def sum(num1, num2):
    return num1 + num2

现在,我把@multiply(5)放在sum()上,然后调用sum(4,6),如下所示:

# (4 + 6) x 5 = 50

@multiply(5) # Here
def sum(num1, num2):
    return num1 + num2

result = sum(4, 6)
print(result)

那么,我可以得到如下结果:

50

接下来,我把@multiply()放在sum()上,然后调用sum(4,6),如下所示:

# (4 + 6) x 1 = 10

@multiply() # Here
def sum(num1, num2):
    return num1 + num2
    
result = sum(4, 6)
print(result)

或者,我把@multiply放在sum()上,然后调用sum(4,6),如下所示:

# 4 + 6 = 10

@multiply # Here
def sum(num1, num2):
    return num1 + num2
    
result = sum(4, 6)
print(result)

那么,我可以得到如下结果:

10

我想展示一个想法,在我看来很优雅。t.dubrownik提出的解决方案显示了一个始终相同的模式:无论装饰器做什么,您都需要三层包装器。

所以我认为这是一个元装饰师的工作,也就是说,装饰师的装饰师。由于decorator是一个函数,它实际上是一个带有参数的常规decorator:

def parametrized(dec):
    def layer(*args, **kwargs):
        def repl(f):
            return dec(f, *args, **kwargs)
        return repl
    return layer

这可以应用于常规的装饰器,以便添加参数。例如,我们有一个decorator,它将一个函数的结果加倍:

def double(f):
    def aux(*xs, **kws):
        return 2 * f(*xs, **kws)
    return aux

@double
def function(a):
    return 10 + a

print function(3)    # Prints 26, namely 2 * (10 + 3)

使用@ parameterized,我们可以构建一个带参数的通用@multiply装饰器

@parametrized
def multiply(f, n):
    def aux(*xs, **kws):
        return n * f(*xs, **kws)
    return aux

@multiply(2)
def function(a):
    return 10 + a

print function(3)    # Prints 26

@multiply(3)
def function_again(a):
    return 10 + a

print function(3)          # Keeps printing 26
print function_again(3)    # Prints 39, namely 3 * (10 + 3)

通常,参数化装饰器的第一个参数是函数,而其余参数将对应于参数化装饰器的参数。

一个有趣的用法示例可以是类型安全的断言装饰器:

import itertools as it

@parametrized
def types(f, *types):
    def rep(*args):
        for a, t, n in zip(args, types, it.count()):
            if type(a) is not t:
                raise TypeError('Value %d has not type %s. %s instead' %
                    (n, t, type(a))
                )
        return f(*args)
    return rep

@types(str, int)  # arg1 is str, arg2 is int
def string_multiply(text, times):
    return text * times

print(string_multiply('hello', 3))    # Prints hellohellohello
print(string_multiply(3, 3))          # Fails miserably with TypeError

最后注意:这里我没有使用functools。包装器函数,但我建议始终使用它。

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(...):
        ...