Django not able to submit form using ModelForm - django

Question

The form is always sending a Get request instead of a Post which has been explicitly added using method = "POST". So, not able to persist the data to db. I have just started with Django so, any help will be appreciated.
Below are the code snippets:
create_order.html
<form method="POST" action="{% url 'home' %}">
{% csrf_token %}
{{form}}
<input class="btn btn-primary btn-sm" type="submit" name="Submit">
</form>
urls.py
urlpatterns = [
path('', views.dashboard, name='home'),
path('products/', views.product, name='products'),
path('customer/<str:cust_id>/', views.customer, name='customer'),
path('create_order/', views.create_order, name='create_order'),
]
views.py
def create_order(request):
form = OrderForm()
if request.method == 'POST':
print("In Post", request.method)
form = OrderForm(request.POST)
if form.is_valid():
form.save()
return redirect('/')
else:
print("In else", request.method)
print(form.is_valid())
if form.is_valid():
form.save()
return redirect('/')
context = {'form' : form}
return render(request, 'accounts/create_order.html', context)
terminal output
In else GET False

Answer 1

You are making a POST request to the wrong view. It should be the create_order view, so:
<form method="POST" action="{% url 'create_order' %}">
{% csrf_token %}
{{ form }}
<input class="btn btn-primary btn-sm" type="submit" name="Submit">
</form>
Note that you should not validate the form in case of a GET request, since then you only render the form. So the view logic should be:
def create_order(request):
if request.method == 'POST':
print("In Post", request.method)
form = OrderForm(request.POST)
if form.is_valid():
form.save()
return redirect('home')
else:
form = OrderForm()
context = {'form' : form}
return render(request, 'accounts/create_order.html', context)

Answer 2

Try to add form action method in your HTML file
<form action="" method="post">{% csrf_token %}
{{ form}}
<input type="submit" value="Submit Feedback" />
</form>
</div>