Django 表单:如何动态创建 ModelChoiceField 标签 [英] Django forms: how to dynamically create ModelChoiceField labels

问题描述

我想为 forms.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:

I would like to create dynamic labels for a forms.ModelChoiceField and I'm wondering how to do that. I have the following form class:

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)

默认标签由配置文件定义中指定的 unicode 函数定义.然而，ModelChoiceField 创建的单选按钮的标签需要动态创建.

The default labels are defined by the unicode function specified in the Profile definition. However the labels for the radio buttons created by the ModelChoiceField need to be created dynamically.

首先，我认为我可以简单地覆盖 Django 文档中描述的 ModelChoiceField.但这会创建静态标签.它允许您定义任何标签，但一旦做出选择，该选择就固定了.

First I thought I could simply override ModelChoiceField as described in the Django documentation. But that creates static labels. It allows you to define any label but once the choice is made, that choice is fixed.

所以我想我需要适应添加一些东西到 init 像:

So I think I need to adapt add something to init like:

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>??? = ???

有人知道如何处理吗?任何帮助将不胜感激.

Anyone having any idea how to handle this? Any help would be appreciated very much.

我找到了一些东西，但我不知道这是否是最好的解决方案.我向 ProfileForm 类的 init 部分添加了一些内容，如下所示:

I found something but I don't know if it's the best solution. I add something to the init part of class ProfileForm as follows:

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)

推荐答案

您可以使用 ModelChoiceField 然后动态更改您的 ProfileForm.__init__ 中的选择，例如 (假设它已经是一个 ModelChoiceField):

You could use a ModelChoiceField and then change the choices in you ProfileForm.__init__ dynamically, eg (assuming that it is already a ModelChoiceField):

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

h.name 在这个例子中将用作选择的标签！

h.name in this example will be used as the label of the choice!

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