django objects.create 方法太慢了如何使更快？

Question

映射了多个表，当我创建发布请求时，

大约需要2~3秒。有什么办法可以解决吗？

我想这需要很长时间：

1. 对象.create

2. for 循环

3. 产品.objects.get

   但是，我无法找到更好的方法..

   楷模：

   #product、Order、OrderItems、ShippingAddress 已映射

   class Order(models.Model):
       user = models.ForeignKey(User, on_delete= models.CASCADE)
       order_date = models.DateTimeField(auto_now=True)
       is_paid = models.BooleanField(default=False)
       paid_at = models.DateTimeField(auto_now=False, null=True, blank=True)
       delivery_code = models.CharField(max_length=255, null=True, blank=True)
       is_delivered = models.BooleanField(default=False)
       delivered_date = models.DateTimeField(auto_now=False, null=True, blank=True)
       total_price = models.DecimalField(max_digits=7, decimal_places=2, null=True)   
       shipping_price = models.DecimalField(max_digits=7, decimal_places=2, null=True)   
       payment_method = models.CharField(max_length=255,null=True) 
       
       def __str__(self):
           return str(self.user)
       
   class OrderItem(models.Model):
       order = models.ForeignKey(Order, on_delete= models.CASCADE, null=True, blank=True)
       product = models.ForeignKey(Product, on_delete= models.CASCADE)
       name = models.CharField(max_length=200, null=True)
       image = models.CharField(max_length=255, null=True)
       qty = models.IntegerField(default=0, null=True)
       price = models.DecimalField(max_digits=7, decimal_places=2, null=True)
       
       def image_preview(self):
           if self.image:
               return mark_safe('<img src="{0}" width="55" height="55" />'.format(self.image))
           else:
               return '(No image)'
       
       def __str__(self):
           return str(self.product)
   
   
   class ShippingAddress(models.Model):
       user = models.ForeignKey(User, on_delete=models.CASCADE)
       order = models.OneToOneField(Order, on_delete=models.CASCADE, null=True, blank=True)
       address = models.CharField(max_length=255, null=False)
       city = models.CharField(max_length=255, null=False)
       postal_code = models.CharField(max_length=255, null=False)
       country = models.CharField(max_length=255, null=False)
       
       def __str__(self):
           return str(self.user)
   

   看法：

   @permission_classes(IsAuthenticated)
   @api_view(['POST'])
   def OrderCreate(request):
       data = request.data
       user = request.user
       order_items = data['orderItems']
       #1.create order
       order = Order.objects.create(
           user = user,
           total_price = data['totalPrice'],
           shipping_price = data['shippingPrice'],
           payment_method = data['paymentMethod']
       )
       
       #2.create orderItems 
       for i in order_items:
           product = Product.objects.get(id=i['id'])
           
           order_item = OrderItem.objects.create(
               order = order,
               product = product,
               name = i['name'],
               qty = i['qty'],
               price = i['price'],
               image = i['image']
           )
           
           #3. update stock
           product.stock -= i['qty']
           product.save()
           
       #4.create shipping address
       
       shipping_address = ShippingAddress.objects.create(
       user = user,
       order = order,
       address = data['shippingAddress']['address'],
       city = data['shippingAddress']['city'],
       postal_code = data['shippingAddress']['postalCode'],
       country = data['shippingAddress']['country'],
           
       )
   
       #5.serializing and save
   
       serializer = OrderSerializer(order, many=False)
       return Response(serializer.data)
   

Answer 1

只要您对 i['id'] 中的产品 ID 有足够的信任，您就可以在不获取产品的情况下实例化 order_items

for i in order_items:
    # product = Product.objects.get(id=i['id'])
    
    order_item = OrderItem.objects.create(
        order = order,
        product_id  = i['id'],   # set the id (magic suffix) without fetching product
        name = i['name'],
        qty = i['qty'],
        price = i['price'],
        image = i['image']
    )

除了使用.create，您可以将这些order_items 实例化为未保存实例的列表并使用OrderItem.bulk_create 创建它们。阅读bulk_create documentation；它有许多警告。

然后，您可以运行一个循环更新产品库存字段，使用 F expression 从产品行中的当前值中减去，而无需实际从数据库中获取产品对象

for i in order_items:
    product_id = i['id']
    Product.objects.filter(
         pk = product_id
    ).update(
         stock = F('stock') - i['qty']
    )

如果您确实将所有产品实例获取到具有更新库存值的列表中，那么还有bulk_update 可以让您在单个数据库操作中应用所有更新的库存值。这可能比用 F 表达式一个一个地做更好。您也可以使用批量获取它们

Product.objects.filter( pk__in=[ i['id'] for i in order_items ] )

（警告，我不认为查询集包含与您提供 i['id'] 值的顺序相同的对象）

将此视为头脑风暴。我不完全确定这是正确的，我真的不知道它是否会加快速度很多，一点点，或者根本没有。我很想知道，如果你尝试一下。