Django: How to take my query out of my for loop - django

Question

I got the following code, that contains N queries:
for qty in total_qty_bought:
product_id = qty["product"]
quantity = int(qty["quantity__sum"])
try:
method_title = (
self.shipment_model.get(order_id=qty["order_id"])
.method_title.replace("Hent-selv", "")
.strip()
)
To solve the issue I tried to take the method_title query out of the for loop like this:
quantity = 0
for qty in total_qty_bought:
quantity = int(qty["quantity__sum"])
method_title = (
self.shipment_model.get(order_id=total_qty_bought[0]['order_id'])
.method_title.replace("Hent-selv", "")
.strip()
)
Note! There will be a full refrence further down, to understand the bigger picture
The issue in my solution is, that I am hard choosing which dict to enter , as I select [0] before order_id, and not in a for loop like before, would be selecting every individual item in the loop.
Is there a more sufficient way to do this? I do not see a solution without the for loop, but django debugger tool tells me it creates 2k+ queries.
CODE FOR REFRENCE
class InventoryStatusView(LoginRequiredMixin, View):
template_name = "lager-status.html"
cinnamon_form = CinnamonForm(prefix="cinnamon_form")
peber_form = PeberForm(prefix="peber_form")
pc_model = InventoryStatus
product_model = Product.objects.all()
order_item_model = WCOrderItem.objects.all()
shipment_model = WCOrderShipment.objects.all()
def get(self, request):
# Get all added objects that hasn't been deleted
objects = self.pc_model.objects.filter(is_deleted=False)
# Get all added objects that has been deleted
deleted_objects = self.pc_model.objects.filter(is_deleted=True)
# Sum all cinnamon that isn't deleted
total_cinnamon = (
self.pc_model.objects.filter(is_deleted=False)
.aggregate(Sum("cinnamon"))
.get("cinnamon__sum", 0.00)
)
# Sum all peber that isn't deleted
total_peber = (
self.pc_model.objects.filter(is_deleted=False)
.aggregate(Sum("peber"))
.get("peber__sum", 0.00)
)
# Get the amount of kilo attached to products
product_data = {}
queryset = ProductSpy.objects.select_related('product')
for productSpy in queryset:
product_data[productSpy.product.product_id] = productSpy.kilo
# Get quantity bought of each product
total_qty_bought = self.order_item_model.values(
"order_id", "product"
).annotate(Sum("quantity"))
# Get the cities from the inventory model
cities = dict(self.pc_model.CITIES)
# Set our total dict for later reference
our_total = {}
product = Product.objects.filter(
product_id__in={qty['product'] for qty in total_qty_bought}
).first()
# Check if we deal with kanel or peber as a product based on slug
index = 0
if product.slug.startswith("kanel-"):
index = 0
elif product.slug.startswith("peber-"):
index = 1
else:
pass
try:
# Sum the total quantity bought
quantity = 0
for qty in total_qty_bought:
quantity = int(qty["quantity__sum"])
# Get the inventory the order is picked from based on shipment method title
method_title = (
self.shipment_model.get(order_id=total_qty_bought[0]['order_id']) # The error
.method_title.replace("Hent-selv", "")
.strip()
)
# If the order isn't picked, but sent, use this inventory
if method_title not in cities.values():
method_title = "Hovedlager"
try:
# Get the total of kanel and peber bought
kilos = quantity * product_data[product.id]
# If there is no inventory, set it to 0 peber and 0 kanel
if method_title not in our_total:
our_total[method_title] = [0, 0]
# Combine place and kilos
our_total[method_title][index] += kilos
except KeyError as ex:
print(ex)
pass
except WCOrderShipment.DoesNotExist as ed:
print(ed)
pass
# Quantities BOUGHT! (in orders!)
print(our_total)
context = {
"cinnamon_form": self.cinnamon_form,
"peber_form": self.peber_form,
"objects": objects,
"deleted_objects": deleted_objects,
"total_cinnamon": total_cinnamon,
"total_peber": total_peber,
"our_total": our_total,
}
return render(request, self.template_name, context)

Answer 1

You can only do one query by using the __in operator:
shipments = self.shipment_model.get(order_id__in=list_containing_order_ids)
Then you can do a normal for loop in which you verify that condition.