Python Django:ValueError:尝试在clean中创建失败时，无法在ForeignField上分配none
我在弄清楚如何在提交表单时自动为ForeignKey字段创建模型实例时遇到了很多麻烦。下面是一个简单的玩具网站，说明了这个问题： 我有两种型号，型号1和型号2。Model2包含Model1的外键。我希望用户能够创建Model2的实例，具体方法是选择要存储在ForeignKey中的Model1实例，或者将该值留空并自动生成Model1实例 我觉得代码应该是这样的。My models.py代码非常简单：Python Django:ValueError:尝试在clean中创建失败时，无法在ForeignField上分配none,python,django,django-forms,Python,Django,Django Forms,我在弄清楚如何在提交表单时自动为ForeignKey字段创建模型实例时遇到了很多麻烦。下面是一个简单的玩具网站，说明了这个问题： 我有两种型号，型号1和型号2。Model2包含Model1的外键。我希望用户能够创建Model2的实例，具体方法是选择要存储在ForeignKey中的Model1实例，或者将该值留空并自动生成Model1实例 我觉得代码应该是这样的。My models.py代码非常简单： # models.py from django.db import models from dj
# models.py
from django.db import models
from django.core.validators import MinValueValidator
class Model1(models.Model):
# Note this field cannot be negative
my_field1 = models.IntegerField(validators=[MinValueValidator(0)])
class Model2(models.Model):
# blank = True will make key_to_model1 not required on the form,
# but since null = False, I will still require the ForeignKey
# to be set in the database.
related_model1 = models.ForeignKey(Model1, blank=True)
# Note this field cannot be negative
my_field2 = models.IntegerField(validators=[MinValueValidator(0)])
#forms.py
from django import forms
from django.forms.models import model_to_dict
from .models import Model1, Model2
# A ModelForm used for validation purposes only.
class Model1Form(forms.ModelForm):
class Meta:
model = Model1
class Model2Form(forms.ModelForm):
class Meta:
model = Model2
def clean(self):
cleaned_data = super(Model2Form, self).clean()
if not cleaned_data.get('related_model1', None):
# Don't instantiate field2 if it doesn't exist.
val = cleaned_data.get('my_field2', None)
if not val:
raise forms.ValidationError("My field must exist")
# Generate a new instance of Model1 based on Model2's data
new_model1 = Model1(my_field1=val)
# validate the Model1 instance with a form form
validation_form_data = model_to_dict(new_model1)
validation_form = Model1Form(validation_form_data)
if not validation_form.is_valid():
raise forms.ValidationError("Could not create a proper instance of Model1.")
# set the model1 instance to the related model and save it to the database.
new_model1.save()
cleaned_data['related_model1'] = new_model1
return cleaned_data
总的来说，我可以想出很多方法来解决这个问题，但我不知道如何以一种合理干净的方式解决这个问题。我找到了一个我认为可以接受的解决方案，这是基于karthikr在评论中的讨论。我肯定仍然对其他选择持开放态度 其思想是使用视图中的逻辑在两种形式之间进行选择以进行验证：一种形式是标准模型形式，另一种是不带ForeignKey字段的模型形式 所以，我的models.py是相同的 My forms.py有两个Model2表单。。。一个非常简单的一个，一个没有ForeignKey字段，使用新逻辑为ForeignKey动态生成Model1的新实例。新表单的干净逻辑就是我在Model2Form中使用的干净逻辑：
#forms.py
from django import forms
from django.forms.models import model_to_dict
from .models import Model1, Model2
# A ModelForm used for validation purposes only.
class Model1Form(forms.ModelForm):
class Meta:
model = Model1
class Model2Form(forms.ModelForm):
class Meta:
model = Model2
# This inherits from Model2Form so that any additional logic that I put in Model2Form
# will apply to it.
class Model2FormPrime(Model2Form):
class Meta:
model = Model2
exclude = ('related_model1',)
def clean(self):
cleaned_data = super(Model2Form, self).clean()
if cleaned_data.get('related_model1', None):
raise Exception('Huh? This should not happen...')
# Don't instantiate field2 if it doesn't exist.
val = cleaned_data.get('my_field2', None)
if not val:
raise forms.ValidationError("My field must exist")
# Generate a new instance of Model1 based on Model2's data
new_model1 = Model1(my_field1=val)
# validate the Model1 instance with a form form
validation_form_data = model_to_dict(new_model1)
validation_form = Model1Form(validation_form_data)
if not validation_form.is_valid():
raise forms.ValidationError("Could not create a proper instance of Model1.")
# set the Model1 instance to the related model and save it to the database.
cleaned_data['related_model1'] = new_model1
return cleaned_data
def save(self, commit=True):
# Best to wait til save is called to save the instance of Model1
# so that instances aren't created when the Model2Form is invalid
self.cleaned_data['related_model1'].save()
# Need to handle saving this way because otherwise related_model1 is excluded
# from the save due to Meta.excludes
instance = super(Model2FormPrime, self).save(False)
instance.related_model1 = self.cleaned_data['related_model1']
instance.save()
return instance
# Create your views here.
from django.views.generic.edit import CreateView
from django.http import HttpResponseRedirect
from .forms import Model2Form, Model2FormPrime
class Model2CreateView(CreateView):
form_class = Model2Form
template_name = 'form_template.html'
success_url = '/add/'
def post(self, request, *args, **kwargs):
if request.POST.get('related_model', None):
# Complete data can just be sent to the standard CreateView form
return super(Model2CreateView, self).post(request, *args, **kwargs)
else:
# super does this, and I won't be calling super.
self.object = None
# use Model2FormPrime to validate the post data without the related model.
validation_form = Model2FormPrime(request.POST)
if validation_form.is_valid():
return self.form_valid(validation_form)
else:
# Create a normal instance of Model2Form to be displayed to the user
# Insantiate it with post data and validation_form's errors
form = Model2Form(request.POST)
form._errors = validation_form._errors
return self.form_invalid(form)
不过这有点难看，因为它需要我使用两个表单来完成一个表单的工作，在表单之间传递错误。因此，如果有人能提出任何建议，我绝对愿意接受其他解决方案。我想你可能至少需要交换保存和清理的数据分配（如model1.save-then-related\u-model1=model1）嗨，杰夫，谢谢你的快速回复！我认为顺序应该无关紧要，因为这些变量是引用，并且模型已经过验证，所以save应该永远不会失败。不过，你点的菜看起来更自然，所以我要换一种。但是，当错误发生时，此代码永远不会到达，因此这不是错误的原因；pdb.在创建新的\u model1之前将\u trace（）设置为，然后您可以一次运行一行（通过使用n）并将新的\u model1的值发送到您的终端