Django model won't update using .update()

Question

I am trying to update a record in my database using info provided by my website's front-end using Model.objects.update(**kwargs) however on running the code I get the following error:

"'Record' with this Id already exists."

I would have expected that using the "update" method instead of the "create" method would mean that it shouldn't matter / should be expected that there is already a record with the same ID.

def find_patient(request): my_form = ExampleForm() if request.method == "POST": my_form = ExampleForm(request.POST) if my_form.is_valid(): UserInfo.objects.update(**my_form.cleaned_data) else: print(my_form.errors) context = { 'form': my_form, 'data_input': DataInput.objects, 'sections': SECTION_CHOICES } return render(request, 'example.html', context)

The resulting error looks tike this:

id

User info with this Id already exists.

Am I misunderstanding the use case for "update" and if so, how would I go about using kwargs to update all fields for a given record in my db?

Your update aims to update all records, not just a single one. Since it includes the primary key, you thus aim to set all the ids to a specific value, which of course raises an error. — willeM_ Van Onsem
– willeM_ Van Onsem, Commented Aug 16, 2019 at 15:48
you are doing it wrong way please check this out stackoverflow.com/questions/4673985/… — Bob White
– Bob White, Commented Aug 16, 2019 at 15:49

willeM_ Van Onsem · Accepted Answer · 2019-08-16 15:55:37Z

Your UserInfo.objects.update(**my_form.cleaned_data) aims to update all records in the database. Since your form includes a primary key (id), it will thus aim to update the id of all records to a given value. Since these values however should be unque, that means it will of course raise an error.

You can prevent that by filtering before the update, like:

UserInfo.objects.filter( pk=my_form.cleaned_data['id'] ).update(**my_form.cleaned_data)

But updating a specific record is usually done differently: you usually pass the primary key in the URL. So the urls.py looks like:

# app/urls.py from django.urls import path from app.views import find_patient urlpatterns = [ # ... path('updatepatient/<int:pk>/', views.find_patient), # ... ]

Then in the view, we can pass the item to update to the form:

# app/views.py from django.shortcuts import get_object_or_404, redirect def find_patient(request, pk): instance = get_object_or_404(UserInfo, pk=pk) if request.method == "POST": my_form = ExampleForm(request.POST, instance=instance) if my_form.is_valid(): my_form.save() return redirect('some-view-name') else: my_form = ExampleForm() context = { 'form': my_form, 'data_input': DataInput.objects.all(), 'sections': SECTION_CHOICES } return render(request, 'example.html', context)

In case of a successful POST request, one usually redirects to a page. This is the well-known Post/Redirect/Get pattern [wiki].

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