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Django错误:为关键字参数获取多个值

django-forms django

Marcus Whybrow · 技术社区 · 15 年前

__init__() got multiple values for keyword argument 'collection_type'

这个 __init__() 函数(如下所示)与本文完全相同,但具有 # code

def __init__(self, collection_type, user=None, parent=None, *args, **kwargs):
    # code
    super(self.__class__, self).__init__(*args, **kwargs)

form = CreateCollectionForm(
    request.POST, 
    collection_type=collection_type, 
    parent=parent, 
    user=request.user
)

编辑:这是构造函数的完整代码

def __init__(self, collection_type, user=None, parent=None, *args, **kwargs):
    self.collection_type = collection_type
    if self.collection_type == 'library':
        self.user = user
    elif self.collection_type == 'bookshelf' or self.collection_type == 'series':
        self.parent = parent
    else:
        raise AssertionError, 'collection_type must be "library", "bookshelf" or "series"'
    super(self.__class__, self).__init__(*args, **kwargs)

编辑:Stacktrace

Environment:

Request Method: POST
Request URL: http://localhost:8000/forms/create_bookshelf/hello
Django Version: 1.1.1
Python Version: 2.6.1
Installed Applications:
['django.contrib.auth',
 'django.contrib.contenttypes',
 'django.contrib.sessions',
 'django.contrib.sites',
 'libraries',
 'users',
 'books',
 'django.contrib.admin',
 'googlehooks',
 'registration']
Installed Middleware:
('django.middleware.common.CommonMiddleware',
 'django.contrib.sessions.middleware.SessionMiddleware',
 'django.contrib.auth.middleware.AuthenticationMiddleware')


Traceback:
File "/Library/Python/2.6/site-packages/django/core/handlers/base.py" in get_response
  92.                 response = callback(request, *callback_args, **callback_kwargs)
File "/Library/Python/2.6/site-packages/django/contrib/auth/decorators.py" in __call__
  78.             return self.view_func(request, *args, **kwargs)
File "/Users/marcus/Sites/marcuswhybrow.net/autolib/libraries/forms.py" in     create_collection
  13.           form = CreateCollectionForm(request.POST,     collection_type=collection_type, user=request.user)

Exception Type: TypeError at /forms/create_bookshelf/hello
Exception Value: __init__() got multiple values for keyword argument 'collection_type'

3 回复 | 直到 15 年前

Daniel Roseman 15 年前

你正在通过考试 collection_type collection_type=collection_type 在对表单构造函数的调用中。因此Python将其包含在 kwargs dictionary-但由于您在该函数的定义中将其声明为位置参数,因此它会尝试传递它两次,从而导致错误。

user=None, parent=None 之前这个 *args 字典,因为它们已经是夸尔格了,并且args必须总是在夸尔格之前。解决方法是删除集合类型、用户和父项的显式定义,并从函数中的kwargs中提取它们:

def __init__(self, *args, **kwargs):
    collection_type = kwargs.pop('collection_type', None)
    user = kwargs.pop('user', None)
    parent = kwargs.pop('parent', None)

gruszczy 15 年前

In [8]: class A:
   ...:     def __init__(self, a, *args):
   ...:         print a, args
   ...:         
   ...:         

In [9]: A(None, a=None)
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)

/home/gruszczy/Programy/logbuilder/<ipython console> in <module>()

TypeError: __init__() got multiple values for keyword argument 'a'

将request.POST移动到调用中的其他位置,但请记住,命名参数位于那些参数之后,而不是。

Ian Clelland 15 年前

Daniel Roseman的解决方案是处理 *args 和 **kwargs

你已经定义了 CreateCollectionForm.__init__ 签名如下:

def __init__(self, collection_type, user=None, parent=None, *args, **kwargs)

form = CreateCollectionForm(
    request.POST, 
    collection_type=collection_type, 
    parent=parent, 
    user=request.user
)

self collection_type 另一个 ,则必须抛出一个TypeError。

super() 给更高层次的建设者。或者,您需要将post dictionary作为 __init__ 方法,并将其传递给超类。