一种方法是使用自定义Select小部件，该小部件允许通过小部件选项的label部分传递选项中的各个属性：（代码来自这个伟大的答案）

class SelectWithOptionAttribute(Select):
   
"""
Use a dict instead of a string for its label. The 'label' key is expected
for the actual label, any other keys will be used as HTML attributes on
the option.
"""

def create_option(self, name, value, label, selected, index, 
                  subindex=None, attrs=None):
    # This allows using strings labels as usual
    if isinstance(label, dict):
        opt_attrs = label.copy()
        label = opt_attrs.pop('label')
    else: 
        opt_attrs = {}
    option_dict = super().create_option(name, value, 
        label, selected, index, subindex=subindex, attrs=attrs)
    for key,val in opt_attrs.items():
        option_dict['attrs'][key] = val
    return option_dict

要填充单个选项，请覆盖ModelChoiceField子类上的label_from_instance方法（参见django docs）：

IngredientChoiceField(ModelChoiceField):
"""ChoiceField with puts ingredient-type on <options>"""

# Use our custom widget:
widget = SelectWithOptionAttribute

def label_from_instance(self, obj):
# 'obj' will be an Ingredient
    return {
        # the usual label:
        'label': super().label_from_instance(obj),
        # the new data attribute:
        'data-ingredient-type': obj.type
    }

最后，在表单中简单使用此字段：

RecipeModelForm(ModelForm):

class Meta:
    model = Recipe
    fields = [
        # other fields ...
        'ingredients',
    ]
    
    field_classes = {
        'ingredients': IngredientChoiceField
    }

为什么不手动渲染字段？
就像

<select>
  {% for option in form.ingredient.choices %}
     <option value="{{ option.id }}" data-ingredient-type={{ option.type }}>{{ option.name }}</option>
  {% endfor %}
</select>

或者可能在你的模型表单类中你给它添加了属性，但是这必须是一个字符串（或者可能是一个函数）

widgets = { ...
     'ingredients' = forms.Select(attrs={'data-ingredient-type': 'fruit'}),
   ...}

我的解决方案是创建一个覆盖create_option()的自定义小部件：

class IngredientSelect(forms.Select):
    def create_option(
        self, name, value, label, selected, index, subindex=None, attrs=None
    ):
        option = super().create_option(
            name, value, label, selected, index, subindex, attrs
        )
        if value:
            ingredient = models.Ingredient.objects.get(pk=value)
            option['attrs'].update({
                'data-type': ingredient.type
            })
        return option

然后，您需要指定用于表单中配料字段的小部件：

class RecipeForm(forms.ModelForm):
    class Meta:
        model = models.Recipe
        fields = '__all__'
        widgets = {
            'ingredient': IngredientSelect,
        }

感谢In Django form, custom SelectField and SelectMultipleField为我指明了这个解决方案。
我对这个解决方案并不完全满意，因为它假设value是Ingredient的pk，并执行直接数据库查询以获取Ingredient选项，似乎模型对象应该可以从ModelChoiceField获得，但我无法找到获取它的方法。

这在新版Django上变得容易多了：

class SelectWithAttributeField(forms.Select):
    def create_option(
        self, name, value, label, selected, index, subindex=None, attrs=None
    ):
        option = super().create_option(
            name, value, label, selected, index, subindex, attrs
        )

        if value:
            option["attrs"]["data-ingredient-type"] = value.instance. ingredient
        return option

在配方模型表单中，将其用作：

class RecipeForm(forms.ModelForm):

    class Meta:
        model = Recipe
        fields = "__all__"
        widgets = {
            "ingredient": SelectWithAttributeField,
        }
    
    def __init__(self, *args, **kwargs):
        super(RecipeForm, self).__init__(*args, **kwargs)

        self.fields["ingredient"].queryset = Ingredient.objects.filter(
            recipe=recipe
        )