Question

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：

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 功能定义。但是，需要动态创建ModelChoiceField创建的单选按钮的标签。

首先，我认为我可以简单地重写ModelChoiceField，如Django文档中所述。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，该选择是固定的。

所以我认为我需要适应添加到 init 的内容，如：

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>??? = ???

任何人都知道如何处理这个问题？非常感谢任何帮助。

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，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

您可以使用ModelChoiceField，然后动态更改ProfileForm.__init__中的选项，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]

此示例中的

h.name将用作选择的标签！

Answer 2

您可以创建自己的表单字段类并覆盖生成标签的方法：

class MyChoiceField(ModelChoiceField):
   def label_from_instance(self, obj):
        # return your own label here...
        return smart_unicode(obj)

以与使用ModelChoiceField相同的形式使用它：

horoscope = MyChoiceField(queryset = .....)

Answer 3

实际上，最后一个代码示例包含错误：

# this function is added
def get_label(obj):
    return '%s: Euro %.2f' % (HoroscopeLanguage.objects.get(horoscope=obj, language__language_code=language_code).horoscope_type_language, obj.price)
.
.
.

self.fields['horoscope'].label_from_instance = get_labels

然后它有效。使用'lambda obj：...'或'def get_label（obj）没有区别：...'

Django表单：如何动态创建ModelChoiceField标签

3 个答案: