Create one-to-one model record, or update if it already exists, in this Django view

Question

UserComputedInfo model has a one-to-one relationship with CustomUser model.
I want to update or create a record in UserComputedInfo during a form update to CustomUser. I managed to create/update a record in views.py

Here is the code;

from .models import UserComputedInfo 

def profile(request, username):
        if request.method == "POST":
            user = request.user
            form = UserUpdateForm(request.POST, request.FILES, instance=user)
            if form.is_valid():
                user_form = form.save()
                # How to save post_code to copy_input in UserComputedInfo model
                user_computed_info_instance = UserComputedInfo(user_id=user.id)
                user_computed_info_instance.copy_input = 1.18 # or whatever value to be assigned
                user_computed_info_instance.save()
    
                messages.success(request, f'{user_form.username}, Your profile has been updated!')
                return redirect("profile", user_form.username)
    
        return redirect("homepage")

This is a follow-up question to a flawed answer I posted to my own past question.
Save values into another database table with one-to-one relationship in Django during form update

The answer works the first time when a new record is being created. However, it fails to work the second time profile() is run because there is already an existing record.

How do I modify profile() such that a new record will be created if there is no existing record and an update will be made if the record already exists?

I am using Django v4, python v3.9, sqlite database on Windows 10

Asked By: user3848207

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Source

Answer 1

You can use Django’s get_or_create Model method, leaving rest of the logics untouched.

 user_computed_info_instance, created  = UserComputedInfo.objects.get_or_create(user_id=user.id)

It’ll create the object first time but will get the existing object next time for the same id.

Answered By: ThePyGuy

Create one-to-one model record, or update if it already exists, in this Django view

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