django内嵌表单与自定义表单
嗨 我有一个在Django应用程序中使用的域模型,我想用一个表单来展示它。我已经用自定义模型表单创建了我的应用程序(没有太多更改,一些字段被排除在外等等)。模型的依赖关系如下所示:django内嵌表单与自定义表单,django,django-forms,Django,Django Forms,嗨 我有一个在Django应用程序中使用的域模型,我想用一个表单来展示它。我已经用自定义模型表单创建了我的应用程序(没有太多更改,一些字段被排除在外等等)。模型的依赖关系如下所示: Complaint \ .--- CarInfo .--- Customer 我的视图函数如下所示: def make(request): if request.method == 'POST': parameters = copy.copy(request.POST) c
Complaint
\
.--- CarInfo
.--- Customer
我的视图函数如下所示:
def make(request):
if request.method == 'POST':
parameters = copy.copy(request.POST)
complaint = Complaint()
carInfo = CarInfo()
customer = Customer()
customer_form = CustomerForm(parameters, instance=customer)
carInfo_form = CarInfoForm(parameters, instance=carInfo)
parameters['complaint_date'] = get_current_date()
parameters['customer'] = 1 # dummy value to allow validation success
parameters['car_info'] = 1 # dummy value to allow validation success
form = ComplaintForm(parameters, instance=complaint)
if form.is_valid() and customer_form.is_valid() and carInfo_form.is_valid():
carInfo_form.save()
customer_form.save()
parameters['customer'] = customer.id
parameters['car_info'] = carInfo.id
form = ComplaintForm(parameters, instance=complaint)
form.save()
return index(request)
else:
form = ComplaintForm()
carInfo_form = CarInfoForm()
customer_form = CustomerForm()
return render_to_response('complaints/make_complaint.html', {'complaint_form' : form, 'customer_form' : customer_form, 'carInfo' : carInfo_form})
complaint_info = form.cleaned_data.get('some_complaint_info')
complaint_instance = Complaint(car_info=car_instance, customer_info=customer_instance, some_complaint_info=some_complaint_info)
complaint_instance.save()
我不太喜欢这种方法,而且它并不适用于所有环境——我还没有找到它不起作用的原因。我一直在研究如何修复这段代码,并发现了类似于内联表单集(http://docs.djangoproject.com/en/dev/topics/forms/modelforms/#inline-格式集)。这个解决方案似乎还可以,但因为我的表单是定制的,所以我可以使用它
也许有人可以给我一些建议,如何妥善解决这种情况。更清洁的解决方案非常受欢迎
已编辑
对我来说,这个解决方案根本不起作用。尽管在外键上设置了伪值,但当我调用is_valid()时,我得到了FALSE,错误消息说这些字段没有设置。我注意到django 1.2.5出现了这个问题-它发生在我打算运行这个应用程序的服务器上,但是我的笔记本电脑(也是django 1.2.5)没有这个问题。我认为你已经有了最干净、最简单的方法,但是如果你想使用表单集,请尝试以下链接:
编辑。我想你可能会因为这个伪值(以及修改request.POST,我可以继续猜测:)而遇到问题,但是@kriegar展示了如何避免这个问题。无论如何,在一个视图中保存多个表单没有什么困难;Django表单足以支持这种情况。我的观点是,明确地这样做是最干净、最简单的方法,表单集不会对情况有太大的改善。您可以将投诉模型的投诉日期更改为类似的内容 投诉日期=models.DateField(默认值=datetime.date.today()) 这样你就可以摆脱
参数['complaint\u date']=get\u current\u date()
对于您的多表单视图,您可以使用未绑定的表单来实现所需的行为
通过将fk排除在投诉表上的车辆和客户之外,投诉表应有效。同时检查所有3个表单的.is_valid(),然后保存投诉对象所依赖的2个表单,创建投诉对象而不提交到数据库(commit=False),将客户和汽车的id添加到该对象,然后保存
在你看来
def make(request):
if request.method == 'POST':
customer_form = CustomerForm(request.POST)
carInfo_form = CarInfoForm(request.POST)
form = ComplaintForm(request.POST)
if form.is_valid() and customer_form.is_valid() and carInfo_form.is_valid():
car_instance = carInfo_form.save()
customer_instance = customer_form.save()
complaint_instance = form.save(commit=False)
complaint_instance.car_info = car_instance
complaint_instance.customer_info = customer_instance
complaint_instance.save()
return index(request)
else:
form = ComplaintForm()
carInfo_form = CarInfoForm()
customer_form = CustomerForm()
context = { 'complaint_form' : form,
'customer_form' : customer_form,
'carInfo' : carInfo_form,
}
return render_to_response('complaints/make_complaint.html', context, context_instance=RequestContext(request))
编辑:
模型如下所示:
class CarInfo(models.Model):
some_car_info = models.CharField()
class Customer(models.Model):
some_customer_info = models.CharField()
class Complaint(models.Model):
car_info = models.ForeignKey(CarInfo)
customer_info = models.ForeignKey(Customer)
some_complaint_info = models.CharField()
class CarInfoForm(forms.ModelForm):
class Meta:
model = CarInfo
class CustomerForm(forms.ModelForm):
class Meta:
model = Customer
class ComplaintForm(forms.ModelForm):
class Meta:
model = Complaint
exclude = ('car_info', 'customer_info',) # or include = ('some_complaint_info',)
forms.py应该如下所示:
class CarInfo(models.Model):
some_car_info = models.CharField()
class Customer(models.Model):
some_customer_info = models.CharField()
class Complaint(models.Model):
car_info = models.ForeignKey(CarInfo)
customer_info = models.ForeignKey(Customer)
some_complaint_info = models.CharField()
class CarInfoForm(forms.ModelForm):
class Meta:
model = CarInfo
class CustomerForm(forms.ModelForm):
class Meta:
model = Customer
class ComplaintForm(forms.ModelForm):
class Meta:
model = Complaint
exclude = ('car_info', 'customer_info',) # or include = ('some_complaint_info',)
让我们浏览一下我在上面写的视图:
- 第一次通过时,没有request.method,所以我们创建了3个表单
- 这些表单将传递到模板并呈现
- 当使用request.method==“POST”求值true再次调用视图时,我们使用request.POST中的数据创建3个绑定表单实例
if request.method == 'POST': customer_form = CustomerForm(request.POST) carInfo_form = CarInfoForm(request.POST) form = ComplaintForm(request.POST)
- 接下来,我们在每个表单上调用.is\u valid()方法。在我们的示例中,因为我们排除了投诉模型表单中的“customer_info”和“car_info”外键字段,所以每个表单只检查字符输入字段是否有效
- 如果验证全部通过,那么我们可以开始将表格保存到模型中,在这里,我们需要小心填充投诉所需的fk:
if form.is_valid() and customer_form.is_valid() and carInfo_form.is_valid(): car_instance = carInfo_form.save() customer_instance = customer_form.save()
- 使用这两个表单,我们可以像往常一样调用.save()。然而,我们将把返回值分配给car_实例和customer_实例。这些将包含我们刚刚使用表单上的.save()方法创建的CarInfo和客户模型的实例
- 接下来,使用.save()方法中的参数,我们可以从绑定表单(包含request.POST数据)创建一个对象,而不是将其保存到数据库中
complaint_instance = form.save(commit=False) complaint_instance.car_info = car_instance complaint_instance.customer_info = customer_instance complaint_instance.save()
- 为了更清楚,您还可以创建一个新的投诉对象,如下所示:
def make(request): if request.method == 'POST': parameters = copy.copy(request.POST) complaint = Complaint() carInfo = CarInfo() customer = Customer() customer_form = CustomerForm(parameters, instance=customer) carInfo_form = CarInfoForm(parameters, instance=carInfo) parameters['complaint_date'] = get_current_date() parameters['customer'] = 1 # dummy value to allow validation success parameters['car_info'] = 1 # dummy value to allow validation success form = ComplaintForm(parameters, instance=complaint) if form.is_valid() and customer_form.is_valid() and carInfo_form.is_valid(): carInfo_form.save() customer_form.save() parameters['customer'] = customer.id parameters['car_info'] = carInfo.id form = ComplaintForm(parameters, instance=complaint) form.save() return index(request) else: form = ComplaintForm() carInfo_form = CarInfoForm() customer_form = CustomerForm() return render_to_response('complaints/make_complaint.html', {'complaint_form' : form, 'customer_form' : customer_form, 'carInfo' : carInfo_form})
complaint_info = form.cleaned_data.get('some_complaint_info') complaint_instance = Complaint(car_info=car_instance, customer_info=customer_instance, some_complaint_info=some_complaint_info) complaint_instance.save()
- 渲染
class CarInfoFrm(forms.ModelForm):
class Meta:
model = CarInfo
fields = (....)
carInfoForm = inlineformset_factory(Complaint, CarInfo, form=carInfoFrm,)
CustomerForm = inlineformset_factory(Complaint, Customer, form=carInfoFrm,)
在你看来:
complaint = Complaint()
carInfo = CarInfo()
customer = Customer()
cus_form = CustomerForm(parameters, instance=complaint)
car_form = CarInfoForm(parameters, instance=complaint)
comp_form = ComplaintForm(parameters, instance=complaint)
if cus_form.is_valid() and ...... :
comp = comp_form.save(commit=False)#do not save it yet
comp.<attr> = "some_value" #you can edit your data before save...
comp.save()
car = car_form(commit=False)
# do as complaint form... edit and save...
当您使用内联表单进行更新时,您将使用父Complaint记录对其进行初始化
car_form = CarInfoForm(parameters, instance=complaint)
所以,car_表单不接受carInfo实例,而是接受投诉实例(这是我第一次回答时的错误,所以我更正了它)。如果它创建了一个新记录,它会自动将其绑定到相关的投诉记录。如果是更新,则只更新您想要的字段
对我来说,最好使用框架的方法,而不是自己编写。通过这样做,您将保证由django进行所有验证检查。这与我的解决方案看起来没有太大区别。如果没有carInfo和customer set的外键,表单如何进行验证?@Marcin Cylke我相信.is_valid()正在检查表单中的字段是否有效。验证表单和验证模型是两件不同的事情@Marcin,
ComplaintForm
应该可以很好地验证,因为这些字段已被exclude
d。试试看。DateField
的更好方法是DateField(auto\u now\u add=True)
——“按照当前的实现,将auto\u now或auto\u now\u add设置为True将导致字段设置为editable=False和blank=True。”,确保这不是一个问题。这不是我写的更复杂的版本吗?你能总结一下你的方法的不同之处吗?我仍然担心,如果不设置fk,表单将无法验证。我将在今天晚些时候尝试此操作。请阅读我对上面的表单和模型验证的评论,似乎还有另一个问题。请张贴您的模型和表格。既然您说这完全可以,那么