在Django项目中将Form函数转换为基于类的视图

Question

我正试图转换我的形式，这是在一个功能布局在一个基于类的视图:

这是我伸出来的。

功能:

def add_business_plan(request):
info = Info.objects.all()
if request.method == 'POST':
form = infoForm(request.POST)
if form.is_valid():
form.save()
business_name = form.cleaned_data.get('businessName')
info_id = form.instance.id
messages.success(request, f'PDF created for {business_name}!, No.({info_id})')
return render(request, 'businessplan/businessplan.html', {'form': form, 'successful_submit': True})
else:
form = infoForm()
print(form.errors)
return render(request, 'businessplan/businessplan.html',
{
'form': form,
'successful_submit': False,
"Info": info
}
)

这里是表单

class infoForm(forms.ModelForm):
class Meta:
model = Info
fields = [
'businessName',
]
widgets = {
'problem_summary': RichTextFormField(),
}

Answer 1

您可以尝试以下操作:

class AddBusinessPlan(View):
template_name = 'businessplan/businessplan.html'
form_class = infoForm
def get(self, request, *args, **kwargs):
form = self.form_class
print(form.errors)
return render(request, template_name, {'form': form})
def post(self, request, *args, **kwargs):
form = self.form_class(request.POST)
if form.is_valid():
form.save()
business_name = form.cleaned_data.get('businessName')
info_id = form.instance.id
messages.success(request, f'PDF created for {business_name}!, No.({info_id})')
return render(request, self.template_name, {'form': form, 'successful_submit': True})
else:
return render(request, self.template_name, {'form': form})