在 python 函数中使用 *args、**kwargs 时“获得关键字参数的多个值”

传递命名参数时request通过**kwargs，我收到一个错误 -

Traceback (most recent call last):
  File "testKwargs.py", line 9, in <module>
    load_strategy(request="myReq", backend="myBackend", redirect_uri=None, *args, **kwargs)
  File "testKwargs.py", line 5, in load_strategy
    get_strategy("backends", "strategy", "storage", *args, **kwargs)
TypeError: get_strategy() got multiple values for keyword argument 'request'

代码在testKwargs.py在下面-

def get_strategy(backends, strategy, storage, request=None, backend=None, *args, **kwargs):
    print request

def load_strategy(*args, **kwargs):
    get_strategy("backends", "strategy", "storage", *args, **kwargs)

args = ([],)
kwargs = {"acess_token":"myAccToken", "id":"myId"}
load_strategy(request="myReq", backend="myBackend", redirect_uri=None, *args, **kwargs)

我期望该键会有一个键值对request in the **kwargs of load_strategy被传递给request参数输入get_stragegy，但情况似乎并非如此。

我想弄清楚我在这里缺少什么。

你正在传递一个extra位置参数：

args = ([],)

该元组中有一个值，即列表对象。它在其他三个参数传递给之后应用get_strategy(), so to request。 Python 看到你调用：

get_strategy("backends", "strategy", "storage", [],
             request="myReq", backend="myBackend", redirect_uri=None, 
             acess_token="myAccToken", id="myId")

并且 4 个位置参数适用于backends, strategy, storage and request分别参数。

如果您打算传递 3 个位置参数，请指定args作为空元组：

args = ()

一切都很好：

>>> def get_strategy(backends, strategy, storage, request=None, backend=None, *args, **kwargs):
...     print request
... 
>>> def load_strategy(*args, **kwargs):
...     get_strategy("backends", "strategy", "storage", *args, **kwargs)
... 
>>> args = ()
>>> kwargs = {"acess_token":"myAccToken", "id":"myId"}
>>> load_strategy(request="myReq", backend="myBackend", redirect_uri=None, *args, **kwargs)
myReq