Python 装饰器函数包装器中的参数

Question

对于装饰器来说还很陌生，这会被认为是糟糕的代码吗？如果是，什么是好的替代品？

import functools
def error_handaler_decorator(func):
    @functools.wraps(func)
    def wrapper(error_message_for_wrapper = None, cont = True, *args, **kwargs):
        try:
            return func(*args, **kwargs)
        except:
            if error_message_for_wraper != None:
                # Report error to user in application specific way
            if cont == True:
                return True

@error_handaler_decorator
def some_func(input_for_func):
    # Do a thing.

@error_handaler_decorator
def some_func_in_a_class(self,input):
    # Do another thing.

some_func(error_message_for_wrapper = something bad happened, input_for_func = some_input)

some_class.some_func_in_a_class(error_message_for_wrapper = something bad happened, cont = False, input_for_func = some_input)

这意味着我在调用装饰函数时必须传递包装器变量，我认为我不能传递args，只能传递kwargs，但它允许我根据什么来定义错误消息我传递给函数，而不是在我定义函数时。

代码有效，（至少与我测试过的一样），但我的 IDE（Visual Studio 代码）非常生气，说：

方法调用中出现意外的关键字参数“error_message_for_wrapper”

我真的很想清理我的代码，我看到的替代方案是try: except: 或with:。 try: except: 让我的代码变得混乱，（至少主观上是这样）。

With. 更好，但我宁愿将我的装饰器作为函数，它更适合项目。

我认为我不能将 with 作为函数。

Answer 1

好的，这取决于我相信您使用的 Python 版本。在 python 3 中你可以这样做：

def error_handler_decorator(func):
    @functools.wraps(func)
    def wrapper(*args, error_message_for_wrapper = None, cont = True, **kwargs):
        try:
            return func(*args, **kwargs)
        except:
            if error_message_for_wrapper is not None:
                # Report error to user in application specific way
            if cont:
                return True
    return wrapper

在 python 2（但也适用于 python 3）中，您可以使用：

def error_handler_decorator(func):
    @functools.wraps(func)
    def wrapper(*args, **kwargs):
        error_message_for_wrapper = kwargs.pop('error_message_for_wrapper', None)
        cont = kwargs.pop('cont', False)
        try:
            return func(*args, **kwargs)
        except:
            if error_message_for_wrapper is not None:
                # Report error to user in application specific way
            if cont:
                return True
    return wrapper

Answer 2

这可能是您应该使用上下文管理器而不是装饰器的情况。

from contextlib import contextmanager


@contextmanager
def handler(msg=None, cont=True):
    try:
        yield
    except Exception:
        if msg is not None:
            print(msg)
        if not cont:
            reraise


with handler("Don't divide by zero!"):
    3/0

print("OK")

会输出

Don't divide by zero!
OK

如果您在调用handler 时设置cont=False，您将看到Don't divide by zero，但随后作为重新引发的异常的回溯会阻止打印OK。

---

回到原点，contextlib 还提供了一种将上下文管理器用作装饰器的方法。不过，您必须在没有 contextmanager 的帮助下定义上下文管理器。

from contextlib import ContextDecorator


class handler(ContextDecorator):
    def __init__(self, msg=None, cont=True):
        self.msg = msg
        self.cont = cont

    # ContextDecorator doesn't provide default definitions,
    # so we have to provide something, even it doesn't really
    # do anything.
    def __enter__(self):
        return self

    def __exit__(self, exc_type, exc_value, tb):
        if exc_value is not None and self.msg is not None:
            print(self.msg)

        # Returning true suppresses any exception
        # that may have been raised in the context. Returning false
        # means the exception is raised as usual.
        return self.cont


# Scolds you, but returns None
@handler("Don't divide by zero")
def some_func(x):
    return 3/x

# Scolds you *and* raises the exception
@handler("Don't divide by zero", cont=False)
def some_other_func(x):
    return 3/x