Question

因为我们需要我们的ModelChoiceFields根据它们的使用位置有不同的（非__unicode__）标签，我认为覆盖ModelChoiceField并使其接受一个额外的参数将是一个明智的想法。被调用而不是label_from_instance。我可以为我们需要的每个实例创建一个子类，但这不是真的干，现在是吗？

我的新ModelChoiceField：

import django.forms


class ModelChoiceField(django.forms.ModelChoiceField):
    """Subclasses Django's ModelChoiceField and adds one parameter, `obj_label`.
    This should be a callable with one argument (the current object) which
    returns a string to use as the label of that object or instance."""

    def __init__(self, obj_label=None, *args, **kwargs):
        super(django.forms.ModelChoiceField, self).__init__(*args, **kwargs)
        self.obj_label = obj_label

    def label_from_instance(self, obj):
        if self.obj_label:
            return self.label(obj)
        return super(django.forms.ModelChoiceField, self).label_from_instance(obj)

似乎简单易行。我只对名为obj_label的参数感兴趣，并将所有其余参数传递给ModelChoiceField的__init__函数。或者我想。 Python现在抱怨__init__获得意外的关键字参数......我这样称呼它：nationality = ModelChoiceField(queryset=Country.objects.all(), obj_label=lambda x: x.name(), empty_label=None)，错误是：

Traceback:
File "/home/patrick/spng/lib/python2.7/site-packages/django/core/handlers/base.py" in get_response
  103.                     resolver_match = resolver.resolve(request.path_info)
File "/home/patrick/spng/lib/python2.7/site-packages/django/core/urlresolvers.py" in resolve
  321.                     sub_match = pattern.resolve(new_path)
File "/home/patrick/spng/lib/python2.7/site-packages/django/core/urlresolvers.py" in resolve
  321.                     sub_match = pattern.resolve(new_path)
File "/home/patrick/spng/lib/python2.7/site-packages/django/core/urlresolvers.py" in resolve
  223.             return ResolverMatch(self.callback, args, kwargs, self.name)
File "/home/patrick/spng/lib/python2.7/site-packages/django/core/urlresolvers.py" in callback
  230.         self._callback = get_callable(self._callback_str)
File "/home/patrick/spng/lib/python2.7/site-packages/django/utils/functional.py" in wrapper
  29.         result = func(*args)
File "/home/patrick/spng/lib/python2.7/site-packages/django/core/urlresolvers.py" in get_callable
  97.             mod = import_module(mod_name)
File "/home/patrick/spng/lib/python2.7/site-packages/django/utils/importlib.py" in import_module
  35.     __import__(name)
File "/home/patrick/spng/src/internship/views.py" in <module>
  40. from resume.views import assemble_dict_list
File "/home/patrick/spng/src/resume/views.py" in <module>
  20. from resume.forms import (HobbyForm, LanguageForm, EducationForm,
File "/home/patrick/spng/src/resume/forms.py" in <module>
  29. class NationalityForm(forms.ModelForm):
File "/home/patrick/spng/src/resume/forms.py" in NationalityForm
  31.     nationality = ModelChoiceField(queryset=Country.objects.all(), obj_label=lambda x: x.nationality(), empty_label=None)
File "/home/patrick/spng/src/stageplaza_ng/fields.py" in __init__
  11.         super(django.forms.ModelChoiceField, self).__init__(*args, **kwargs)
File "/home/patrick/spng/lib/python2.7/site-packages/django/forms/fields.py" in __init__
  672.                                         initial=initial, help_text=help_text, *args, **kwargs)

Exception Type: TypeError at /stagiaires/overzicht/
Exception Value: __init__() got an unexpected keyword argument 'queryset'

这可能是非常愚蠢的事情，所以有人有想法吗？

Answer 1

重命名除ModelChoiceField以外的类，请注意剩下的逻辑

 class CustomModelChoiceField(django.forms.ModelChoiceField):
        """Subclasses Django's ModelChoiceField and adds one parameter, `obj_label`.
        This should be a callable with one argument (the current object) which
        returns a string to use as the label of that object or instance."""

        def __init__(self, obj_label=None, *args, **kwargs):
            super(CustomModelChoiceField, self).__init__(*args, **kwargs)
            self.obj_label = obj_label

        def label_from_instance(self, obj):
            if self.obj_label:
                return self.label(obj)
            return super(CustomModelChoiceField, self).label_from_instance(obj)

Answer 2

事实证明这真的很愚蠢。我在超级电话中调用了Django ModelChoiceField super。像这样：super(django.forms.ModelChoiceField, self).__init__(*args, **kwargs)。在这种情况下，我得到这个例外是完全合理的。 ModelChoiceField super是ChoiceField。该类不接受名称为queryset的关键字参数。

事实上，我应该称之为当前班级的超级。像这样：super(ModelChoiceField, self).__init__(*args, **kwargs)。这是文档中的很多次，但这次我有点想念它。 ModelChoiceField的正确覆盖是：

import django.forms


class ModelChoiceField(django.forms.ModelChoiceField):
    """Subclasses Django's ModelChoiceField and adds one parameter, `obj_label`.
    This should be a callable with one argument (the current object) which
    returns a string to use as the label of that object or instance."""

    def __init__(self, obj_label=None, *args, **kwargs):
        super(ModelChoiceField, self).__init__(*args, **kwargs)
        self.obj_label = obj_label

    def label_from_instance(self, obj):
        if self.obj_label:
            return self.label(obj)
        return super(ModelChoiceField, self).label_from_instance(obj)

这很有效。但我不得不部分同意斯里尼瓦斯的回答。重命名为ModelChoiceField更清楚，因此可以立即清楚发生了什么。

覆盖Django的ModelChoiceField

2 个答案: