How to clone a model instance without cloning database relationship? - django

Question

I'm building a cart model with the following code.
from django.db import models
class Item(models.Model):
name = models.CharField(max_length=200)
price = models.DecimalField(max_digits=8, decimal_places=2)
def __str__(self):
return self.name
class Order(models.Model):
date = models.DateTimeField(auto_now_add=True)
transcation_id = models.CharField(max_length=200, null=True)
def __str__(self):
return str(self.date)
class OrderItem(models.Model):
item = models.ForeignKey(Item, on_delete=models.CASCADE, blank=True, null=True)
order = models.ForeignKey(Order, on_delete=models.CASCADE, blank=True, null=True)
quantity = models.IntegerField(default=0, blank=True, null=True)
The many-to-one relationship between Item and Order allows one Order to contain many Item and that looks fine.
A model instance can be simply cloned as already answer in this question.
My problem is, if the price of an Item is changed. The price of contained items in Order is change too. But I don't want it to be changed. In the situation that customer already make a purchase, the price cannot be change. Is there anyway to clone the Order instance that completely not related to the other model?

Answer 1

Save the price manually
class OrderItem(models.Model):
item = models.ForeignKey(Item, on_delete=models.CASCADE, blank=True, null=True)
order = models.ForeignKey(Order, on_delete=models.CASCADE, blank=True, null=True)
quantity = models.IntegerField(default=0, blank=True, null=True)
price = price = models.DecimalField(max_digits=8, decimal_places=2, null=True, default=None)
def save():
if self.pk == None:
self.price = self.item.price
super(OrderItem, self).save()