[Fixed]-Is there a simpler way of creating a Django object in a view with POST parameters..

by

0👍

✅

You can use Python’s kwargs expansion using ** to do this:

enquiry = Enquiry(**kwargs)

But don’t. This is unsafe for the same reason Rails introduced strong_params. Use a form instead.

👤Daniel Roseman

1👍

The correct and the easiest way is to use model forms.

class EnquiryForm(ModelForm):
    class Meta:
        model = Enquiry
        fields = ['name', 'email', 'enquiry_input', 'enquiry_type']

You don’t have to display it to the user, you may just use it for validation and saving:

def my_view(request):
    form = EnquiryForm(request.POST)
    if form.is_valid():
        form.save()
        return # normally a redirect response here
    else:
        # react to incorrect data
        # you may investigate form.errors
        return # some error response
👤Piotr Ćwiek

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