Question

如果我有两种形式，基于不同的基类（比如Form和ModelForm），但我想在两者中使用几个字段，我可以以干燥的方式重用它们吗？

考虑以下情况：

class AfricanSwallowForm(forms.ModelForm):
    airspeed_velocity = forms.IntegerField(some_important_details_here)
    is_migratory = forms.BooleanField(more_important_details)

    class Meta:
        model = AfricanBird

class EuropeanSwallowForm(forms.Form):
    airspeed_velocity = forms.IntegerField(some_important_details_here)
    is_migratory = forms.BooleanField(more_important_details)

....有没有办法可以重用字段airspeed_velocity和is_migratory？想象一下，我有几十种这样的形式。如果我一遍又一遍地写这些代码，代码将会浸泡。

（假设，为了这个问题的目的，我不能或不会将airspeed_velocity和is_migratory转换为模型AfricanBird的字段。）

Answer 1

您可以使用多重继承aka mixins 来分解Form和ModelForm中使用的字段。

class SwallowFormFields:
    airspeed_velocity = forms.IntegerField( ... )
    is_migratory = forms.BooleanField( ... )

class AfricanSwallowForm(forms.ModelForm, SwallowFormFields):
    class Meta:
        model = AfricanBird

class EuropeanSwallowForm(forms.Form, SwallowFormFields):
    pass

<强>更新

由于这不适用于Django元编程，您需要创建一个自定义__init__构造函数，将继承的字段添加到对象的字段列表中，或者您可以在类定义中显式添加引用：

class SwallowFormFields:
    airspeed_velocity = forms.IntegerField()
    is_migratory = forms.BooleanField()

class AfricanSwallowForm(forms.ModelForm):
    airspeed_velocity = SwallowFormFields.airspeed_velocity
    is_migratory = SwallowFormFields.is_migratory
    class Meta:
        model = AfricanSwallow

class EuropeanSwallowForm(forms.Form):
    airspeed_velocity = SwallowFormFields.airspeed_velocity
    is_migratory = SwallowFormFields.is_migratory

<强>更新：

当然，您不必将共享字段嵌套到类中 - 您也可以将它们简单地定义为全局...

airspeed_velocity = forms.IntegerField()
is_migratory = forms.BooleanField()

class AfricanSwallowForm(forms.ModelForm):
    airspeed_velocity = airspeed_velocity
    is_migratory = is_migratory
    class Meta:
        model = AfricanSwallow

class EuropeanSwallowForm(forms.Form):
    airspeed_velocity = airspeed_velocity
    is_migratory = is_migratory

<强>更新

好的，如果你真的想干到最大，你必须使用元类。

所以你可以这样做：

from django.forms.models import ModelForm, ModelFormMetaclass
from django.forms.forms import get_declared_fields, DeclarativeFieldsMetaclass
from django.utils.copycompat import deepcopy

class MixinFormMetaclass(ModelFormMetaclass, DeclarativeFieldsMetaclass):
    def __new__(cls, name, bases, attrs):

        # default __init__ that calls all base classes
        def init_all(self, *args, **kwargs):
            for base in bases:
                super(base, self).__init__(*args, **kwargs)
        attrs.setdefault('__init__', init_all)

        # collect declared fields
        attrs['declared_fields'] = get_declared_fields(bases, attrs, False)

        # create the class
        new_cls = super(MixinFormMetaclass, cls).__new__(cls, name, bases, attrs)
        return new_cls

class MixinForm(object):
    __metaclass__ = MixinFormMetaclass
    def __init__(self, *args, **kwargs):
        self.fields = deepcopy(self.declared_fields)

您现在可以从MixinForm派生您的表单域集合，如下所示：

class SwallowFormFields(MixinForm):
    airspeed_velocity = forms.IntegerField()
    is_migratory = forms.BooleanField()

class MoreFormFields(MixinForm):
    is_endangered = forms.BooleanField()

然后将它们添加到基类列表中，如下所示：

class EuropeanSwallowForm(forms.Form, SwallowFormFields, MoreFormFields):
    pass

class AfricanSwallowForm(forms.ModelForm, SwallowFormFields):
    class Meta:
        model = AfricanSwallow

那它做了什么？

元类收集MixinForm
然后添加自定义__init__构造函数，以确保MixinForm的__init__方法被神奇地调用。（否则你必须明确地调用它。）
MixinForm.__init__将声明的字段复制到字段属性

请注意，我既不是Python大师也不是django开发人员，而且元类很危险。因此，如果您遇到奇怪的行为，请更好地坚持以上更详细的方法：）

祝你好运！

Answer 2

工厂式方法怎么样？

def form_factory(class_name, base, field_dict):
    always_has = {
        'airspeed_velocity': forms.IntegerField(some_important_details_here),
        'is_migratory': forms.BooleanField(more_important_details)
    }
    always_has.update(field_dict)
    return type(class_name, (base,), always_has)

def meta_factory(form_model):
    class Meta:
        model = form_model
    return Meta

AfricanSwallowForm = form_factory('AfricanSwallowForm', forms.ModelForm, {
        'other' = forms.IntegerField(some_important_details_here),
        'Meta': meta_factory(AfricanBird),
    })

EuropeanSwallowForm = form_factory('EuropeanSwallowForm', forms.Form, {
        'and_a_different' = forms.IntegerField(some_important_details_here),
    })

就此而言，您可以在此处修改工厂函数以查看现有的表单类并选择所需的属性，这样您就不会丢失声明性语法......

Answer 3

<击>

class SwallowForm(forms.Form):
    airspeed_velocity = forms.IntegerField()
    is_migratory = forms.BooleanField()

class AfricanSwallowForm(forms.ModelForm, SwallowForm):
    class Meta:
        model = AfricanSwallow

class EuropeanSwallowForm(forms.Form, SwallowForm):
    ...

~~应该有效。~~

我有一些运行良好的代码，并且字段attr看起来像这样。

languages_field = forms.ModelMultipleChoiceField( queryset=Language.objects.all(), widget=forms.CheckboxSelectMultiple, required=False ) class FooLanguagesForm(forms.ModelForm): languages = languages_field class Meta: model = Foo fields = ('languages', )

~~请注意，我仍然使用Meta中的字段元组。~~该字段显示在表单实例的字段中，无论它是否在Meta.fields中。

我有一个使用此模式的常规表单：

genres_field = forms.ModelMultipleChoiceField( queryset=blah, widget=forms.CheckboxSelectMultiple, #required=False, ) class FooGenresForm(forms.Form): genres = genres_field

我看到我的田地字典正在运作。

In [6]: f = FooLanguagesForm() In [7]: f.fields Out[7]: {'languages': <django.forms.models.ModelMultipleChoiceField object at 0x1024be450>} In [8]: f2 = FooGenresForm() In [9]: f2.fields Out[9]: {'genres': <django.forms.models.ModelMultipleChoiceField object at 0x1024be3d0>}

Answer 4

创建IntegerField的子类

class AirspeedField(forms.IntegerField):
    def __init__():
        super(AirspeedField, self).__init__(some_important_details_here)

Answer 5

我刚刚制作了一个解决这个问题的片段：

https://djangosnippets.org/snippets/10523/

它使用了酥脆的形状，但可以使用相同的想法而没有酥脆的形式。我们的想法是在同一个表单标签下使用多个表单。

Django：重用表单字段而不继承？

5 个答案: