Django表单：如何动态创建ModelChoiceField标签

Question

我想为form.ModelChoiceField创建动态标签，我想知道如何做到这一点。我有以下表单类：

class ProfileForm(forms.ModelForm):

    def __init__(self, data=None, ..., language_code='en', family_name_label='Family name', horoscope_label='Horoscope type', *args, **kwargs):
        super(ProfileForm, self).__init__(data, *args, **kwargs)

        self.fields['family_name'].label = family_name_label
        .
        .
        self.fields['horoscope'].label = horoscope_label
        self.fields['horoscope'].queryset = Horoscope.objects.all()

    class Meta:
        model = Profile

    family_name = forms.CharField(widget=forms.TextInput(attrs={'size':'80', 'class': 'contact_form'}))
    .
    .
    horoscope = forms.ModelChoiceField(queryset = Horoscope.objects.none(), widget=forms.RadioSelect(), empty_label=None)

默认标签由Profile定义中指定的unicode函数定义。但是，需要动态创建由ModelChoiceField创建的单选按钮的标签。

首先，我认为我可以按照Django文档中的描述简单地覆盖ModelChoiceField。但这会创建静态标签。它允许你定义任何标签，但是一旦做出选择，该选择就是固定的。

所以我认为我需要适应添加一些东西到init：

class ProfileForm(forms.ModelForm):

    def __init__(self, data=None, ..., language_code='en', family_name_label='Family name', horoscope_label='Horoscope type', *args, **kwargs):
        super(ProfileForm, self).__init__(data, *args, **kwargs)

        self.fields['family_name'].label = family_name_label
        .
        .
        self.fields['horoscope'].label = horoscope_label
        self.fields['horoscope'].queryset = Horoscope.objects.all()
        self.fields['horoscope'].<WHAT>??? = ???

任何人有任何想法如何处理？任何帮助将不胜感激。

我找到了一些东西，但我不知道这是否是最佳解决方案。我将一些内容添加到类ProfileForm 的init部分中，如下所示：

class ProfileForm((forms.ModelForm):

    def __init__(self, data=None, ..., language_code='en', family_name_label='Family name', horoscope_label='Horoscope type', *args, **kwargs):
    super(ProfileForm, self).__init__(data, *args, **kwargs)

        # this function is added
        def get_label(self, language_code):
            """
            returns the label in the designated language, from a related object (table)
            """
            return HoroscopeLanguage.objects.get(horoscope=obj, language__language_code=language_code).horoscope_type_language

        self.fields['family_name'].label = family_name_label
        .
        .
        self.fields['horoscope'].queryset = Horoscope.objects.all()
        self.fields['horoscope'].label_from_instance = lambda obj: "%s: Euro %.2f" % (HoroscopeLanguage.objects.get(horoscope=obj, language__language_code=language_code).horoscope_type_language, obj.price)
        .
        .
        """
        The next code also works, the lambda function without the get_label function
        """
        self.fields['horoscope'].label_from_instance = lambda obj: "%s: Euro %.2f" % (obj.horoscope_type, obj.price)
        .
        .
        """
        But this code doesn't work. Anyone?
        """
        self.fields['horoscope'].label_from_instance = get_label(obj, language_code)

Answer 1

你可以使用a ModelChoiceField然后ProfileForm.__init__动态更改你的选择，例如（假设它已经是ModelChoiceField）：

horoscopes = Horoscope.objects.all()
self.fields['horoscope'].choices = [(h.pk, h.name) for h in horoscopes]

h.name 在本例中将用作选择的标签！