是否可以在django中的单个ModelForm
中包含多个模型?我正在尝试创建个人资料编辑表单。所以我需要在用户模型和 UserProfile模型中包含一些字段。目前我正在使用2个这样的表格
class UserEditForm(ModelForm):
class Meta:
model = User
fields = ("first_name", "last_name")
class UserProfileForm(ModelForm):
class Meta:
model = UserProfile
fields = ("middle_name", "home_phone", "work_phone", "cell_phone")
有没有办法将这些整合到一个表单中,或者我只是需要创建一个表单并处理数据库加载并保存自己?
答案 0 :(得分:82)
您可以在一个<form>
html元素的模板中显示两个表单。然后在视图中单独处理表单。您仍然可以使用form.save()
,而不必处理数据库加载并自行保存。
在这种情况下,您不应该需要它,但如果您要使用具有相同字段名称的表单,请查看prefix
kwarg以获取django表单。 (我回答了一个关于它的问题here)。
答案 1 :(得分:8)
您可以尝试使用这段代码:
class CombinedFormBase(forms.Form):
form_classes = []
def __init__(self, *args, **kwargs):
super(CombinedFormBase, self).__init__(*args, **kwargs)
for f in self.form_classes:
name = f.__name__.lower()
setattr(self, name, f(*args, **kwargs))
form = getattr(self, name)
self.fields.update(form.fields)
self.initial.update(form.initial)
def is_valid(self):
isValid = True
for f in self.form_classes:
name = f.__name__.lower()
form = getattr(self, name)
if not form.is_valid():
isValid = False
# is_valid will trigger clean method
# so it should be called after all other forms is_valid are called
# otherwise clean_data will be empty
if not super(CombinedFormBase, self).is_valid() :
isValid = False
for f in self.form_classes:
name = f.__name__.lower()
form = getattr(self, name)
self.errors.update(form.errors)
return isValid
def clean(self):
cleaned_data = super(CombinedFormBase, self).clean()
for f in self.form_classes:
name = f.__name__.lower()
form = getattr(self, name)
cleaned_data.update(form.cleaned_data)
return cleaned_data
示例用法:
class ConsumerRegistrationForm(CombinedFormBase):
form_classes = [RegistrationForm, ConsumerProfileForm]
class RegisterView(FormView):
template_name = "register.html"
form_class = ConsumerRegistrationForm
def form_valid(self, form):
# some actions...
return redirect(self.get_success_url())
答案 2 :(得分:3)
我写了一个Mixin,因此您可以使用所有通用的基于类的视图。定义模型,字段,现在还有child_model和child_field - 然后你可以在Zach描述的标签中包装两个模型的字段。
class ChildModelFormMixin:
''' extends ModelFormMixin with the ability to include ChildModelForm '''
child_model = ""
child_fields = ()
child_form_class = None
def get_child_model(self):
return self.child_model
def get_child_fields(self):
return self.child_fields
def get_child_form(self):
if not self.child_form_class:
self.child_form_class = model_forms.modelform_factory(self.get_child_model(), fields=self.get_child_fields())
return self.child_form_class(**self.get_form_kwargs())
def get_context_data(self, **kwargs):
if 'child_form' not in kwargs:
kwargs['child_form'] = self.get_child_form()
return super().get_context_data(**kwargs)
def post(self, request, *args, **kwargs):
form = self.get_form()
child_form = self.get_child_form()
# check if both forms are valid
form_valid = form.is_valid()
child_form_valid = child_form.is_valid()
if form_valid and child_form_valid:
return self.form_valid(form, child_form)
else:
return self.form_invalid(form)
def form_valid(self, form, child_form):
self.object = form.save()
save_child_form = child_form.save(commit=False)
save_child_form.course_key = self.object
save_child_form.save()
return HttpResponseRedirect(self.get_success_url())
示例用法:
class ConsumerRegistrationUpdateView(UpdateView):
model = Registration
fields = ('firstname', 'lastname',)
child_model = ConsumerProfile
child_fields = ('payment_token', 'cart',)
或使用ModelFormClass:
class ConsumerRegistrationUpdateView(UpdateView):
model = Registration
fields = ('firstname', 'lastname',)
child_model = ConsumerProfile
child_form_class = ConsumerProfileForm
完成。希望有所帮助。
答案 3 :(得分:2)
你可能应该看看Inline formsets。当您的模型通过外键关联时,将使用内联表单集。
答案 4 :(得分:2)
您可以检查my answer here是否存在类似问题。
它讨论了如何将注册和用户配置文件组合成一个表单,但它可以推广到任何ModelForm组合。
答案 5 :(得分:1)
我在项目中使用了django betterforms的MultiForm and MultiModelForm。但是,可以改进代码。例如,它依赖于django.six,而3. +不支持django.six,但是所有这些都可以轻松修复