models.py:
class Ingredient(models.Model):
_est_param = None
param = models.ManyToManyField(Establishment, blank=True, null=True, related_name='+', through='IngredientParam')
def est_param(self, establishment):
if not self._est_param:
self._est_param, created = self.ingredientparam_set\
.get_or_create(establishment=establishment)
return self._est_param
class IngredientParam(models.Model):
#ingredient params
active = models.BooleanField(default=False)
ingredient = models.ForeignKey(Ingredient)
establishment = models.ForeignKey(Establishment)
I need to fetch all Ingredient with parametrs for Establishment. First I fetch Ingredients.objects.all() and use all params like Ingredients.objects.all()[0].est_param(establishment).active. How I can use django 1.4 prefetch_related to make less sql queries? May be I can use other way to store individual Establishment properties for Ingredient?
Django 1.7 adds the Prefetch object you can put into prefetch_related. It allows you to specify a queryset which should provide the filtering. I'm having some problems with it at the moment for getting a singular (latest) entry from a list, but it seems to work very well when trying to get all the related entries.
You could also checkout django-prefetch which is part of this question which does not seem a duplicate of this question because of the vastly different wording.
The following code would fetch all the ingredients and their parameters in 2 queries:
ingredients = Ingredients.objects.all().prefetch_related('ingredientparam_set')
You could then access the parameters you're interested in without further database queries.
Related
I have a simple question but multiple google searches left me without a nice solution. Currently I am doing the following:
allowed_categories = self.allowed_view.all().difference(self.not_allowed_view.all())
users = []
for cat in allowed_categories:
for member in cat.members.all():
users.append(member)
return users
I have a ManyToManyField to Objects that also have a ManyToManyField for instances of Users. In the code above, I am trying to get all the users from all those categories and get a list of all Users.
Later I would like the same in a method allowed_to_view(self, user_instance) but that's for later.
How would I achieve this using Django ORM without using nested for-loops?
[edit]
My models are as follows:
class RestrictedView(models.Model):
allowed_view = models.ManyToManyField(Category)
not_allowed_view = models.ManyToManyField(Category)
class Category(models.Model):
name = models.CharField(max_length=30)
members = models.ManyToManyField(User)
So, I've made the following one-liner with only one query towards the database. It took me some time...
users = User.objects.filter(pk__in=self.allowed_view.all().values("users").difference(self.not_allowed_view.all().values("users")))
This gives me a nice queryset with only the users that are in the allowed_view and explicitly not in the not_allowed_view.
Without seeing you database structure / models.py file its hard to say, but you can do a search on member objects like so:
members_queryset = Member.objects.filter(
category = <allowed categories>,
...
)
users += list(members.all())
I have 2 models with a one-to-many relation on a MySQL DB:
class Domains(models.Model):
id = models.AutoField(primary_key=True)
name = models.CharField(max_length=50, unique=True)
description = models.TextField(blank=True, null=True)
class Kpis(models.Model):
id = models.AutoField(primary_key=True)
name = models.CharField(max_length=50, unique=True)
description = models.TextField(blank=True, null=True)
domain_id = models.ForeignKey(Domains, on_delete=models.CASCADE, db_column='domain_id')
In order to bring ALL the domains with all their kpis objects, i use this code with a for loop:
final_list = []
domains_list = Domains.objects.all()
for domain in domains_list:
# For each domain, get all related KPIs
domain_kpis = domain.kpis_set.values()
final_list.append({domain:domains_kpis})
The total number of queries i run is: 1 + the number of total domains i have, which is quite a lot.
I'm looking for a way to optimize this, preferably to execute it within only one query on the database. Is this possible?
You use .prefetch_related(…) [Django-doc] for this:
final_list = []
domains_list = Domains.objects.prefetch_related('kpis_set')
for domain in domains_list:
# For each domain, get all related KPIs
domain_kpis = domain.kpis_set.all()
final_list.append({domain:domains_kpis})
This will make two queries: one to query the domains, and a second to query all the related Kpis with a single query into memory.
Furthermore please do not use .values(). You can serialze data to JSON with Django's serializer framework, by making use of .values() you "erode" the model layer. See the Serializing Django objects section of the documentation for more information.
Just wanted to add that you are asking a solution for "classic" N +1 queries problem. Here you can read a something about it and aslo find the examples for prefetch_related method adviced in Willem's answer.
Another thing worth mentioning is that probably you aren't suppose to use this dict final_list.append({domain:domains_kpis}), but instead you may want to map some field(s) from Domain to some field(s) from Kapis models and, if this is true, you can specify exact fields you'd like to have prefetched using Prefetch:
domains_list = Domains.objects.prefetch_related(Prefetch('kpis_set'), queryset=Kapis.objects.all().only('some_field_you_want_to_have'))
final_list = []
for domain in domains_list:
domain_kpis = domain.kpis_set.all()
final_list.append({domain.some_field:domains_kpis.prefetched_field})
This should give another boost to performance on big-volume table's.
I am facing one issue while filtering the data .
I am having three models ...
class Product:
size = models.CharField(max_length=200)
class Make(models.Model):
name = models.ForeignKey(Product, related_name='product_set')
class MakeContent(models.Model):
make = models.ForeignKey(Make, related_name='make_set')
published = models.BooleanField()
I can generate a queryset that contains all Makes and each one's related MakeContents where published = True.
Make.objects.filter(make_set__published=True)
I'd like to know if it's possible (without writing SQL directly) for me to generate a queryset that contains all Product and each one's related MakeContents where published = True.
I have tried this
Product.objects.filter(product_set__make_set__published=True)
But it's not working
A subquery can solve the problem.
from django.db.models import Subquery
sub_query = MakeContent.objects.filter(published=True)
Product.objects.filter(
pk__in=Subquery(sub_query.values('make__name__pk'))
)
Use this
Product.objects.filter(make__makecontent__published=True).distinct()
Filtering through ForeignKey works both forward and backwards in Django. You can use the model name in lower case for backward filtering.
https://docs.djangoproject.com/en/3.1/topics/db/queries/#lookups-that-span-relationships
Finally .distinct() is required to remove the multiple values produced due to SQL joins
Is there a way in Django to write a query using the ORM, not raw SQL that allows you to JOIN on another table without there being a foreign key? Looking through the documentation it appears in order for the One to One relationship to work there must be a foreign key present?
In the models below I want to run a query with a JOIN on UserActivity.request_url to UserActivityLink.url.
class UserActivity(models.Model):
id = models.IntegerField(primary_key=True)
last_activity_ip = models.CharField(max_length=45L, blank=True)
last_activity_browser = models.CharField(max_length=255L, blank=True)
last_activity_date = models.DateTimeField(auto_now_add=True)
request_url = models.CharField(max_length=255L, blank=True)
session_id = models.CharField(max_length=255L)
users_id = models.IntegerField()
class Meta:
db_table = 'user_activity'
class UserActivityLink(models.Model):
id = models.IntegerField(primary_key=True)
url = models.CharField(max_length=255L, blank=True)
url_description = models.CharField(max_length=255L, blank=True)
type = models.CharField(max_length=45L, blank=True)
class Meta:
db_table = 'user_activity_link'
The link table has a more descriptive translation of given URLs in the system, this is needed for some reporting the system will generate.
I've tried creating the foreign key from UserActivity.request_url to UserActivityLink.url but it fails with the following error: ERROR 1452: Cannot add or update a child row: a foreign key constraint fails
No, there isn't an effective way unfortunately.
The .raw() is there for this exact thing. Even if it could it probably would be a lot slower than raw SQL.
There is a blogpost here detailing how to do it with query.join() but as they themselves point out. It's not best practice.
Just reposting some related answer, so everyone could see it.
Taken from here: Most efficient way to use the django ORM when comparing elements from two lists
First problem: joining unrelated models
I'm assuming that your Model1 and Model2 are not related,
otherwise you'd be able to use Django's related objects
interface. Here are two approaches you could take:
Use extra and a SQL subquery:
Model1.objects.extra(where = ['field in (SELECT field from myapp_model2 WHERE ...)'])
Subqueries are not handled very efficiently in some databases
(notably MySQL) so this is probably not as good as #2 below.
Use a raw SQL query:
Model1.objects.raw('''SELECT * from myapp_model1
INNER JOIN myapp_model2
ON myapp_model1.field = myapp_model2.field
AND ...''')
Second problem: enumerating the result
Two approaches:
You can enumerate a query set in Python using the built-in enumerate function:
enumerate(Model1.objects.all())
You can use the technique described in this answer to do the enumeration in MySQL. Something like this:
Model1.objects.raw('''SELECT *, #row := #row + 1 AS row
FROM myapp_model1
JOIN (SELECT #row := 0) rowtable
INNER JOIN myapp_model2
ON myapp_model1.field = myapp_model2.field
AND ...''')
The Django ForeignKey is different from SQL ForeignKey. Django ForeignKey just represent a relation, it can specify whether to use database constraints.
Try this:
request_url = models.ForeignKey(UserActivityLink, to_field='url_description', null=True, on_delete=models.SET_NULL, db_constraint=False)
Note that the db_constraint=False is required, without it Django will build a SQL like:
ALTER TABLE `user_activity` ADD CONSTRAINT `xxx` FOREIGN KEY (`request_url`) REFERENCES `user_activity_link` (`url_description`);"
I met the same problem, after a lot of research, I found the above method.
Hope it helps.
this is a model of the view table.
class QryDescChar(models.Model):
iid_id = models.IntegerField()
cid_id = models.IntegerField()
cs = models.CharField(max_length=10)
cid = models.IntegerField()
charname = models.CharField(max_length=50)
class Meta:
db_table = u'qry_desc_char'
this is the SQL i use to create the table
CREATE VIEW qry_desc_char as
SELECT
tbl_desc.iid_id,
tbl_desc.cid_id,
tbl_desc.cs,
tbl_char.cid,
tbl_char.charname
FROM tbl_desC,tbl_char
WHERE tbl_desc.cid_id = tbl_char.cid;
i dont know if i need a function in models or views or both. i want to get a list of objects from that database to display it. This might be easy but im new at Django and python so i having some problems
Django 1.1 brought in a new feature that you might find useful. You should be able to do something like:
class QryDescChar(models.Model):
iid_id = models.IntegerField()
cid_id = models.IntegerField()
cs = models.CharField(max_length=10)
cid = models.IntegerField()
charname = models.CharField(max_length=50)
class Meta:
db_table = u'qry_desc_char'
managed = False
The documentation for the managed Meta class option is here. A relevant quote:
If False, no database table creation
or deletion operations will be
performed for this model. This is
useful if the model represents an
existing table or a database view that
has been created by some other means.
This is the only difference when
managed is False. All other aspects of
model handling are exactly the same as
normal.
Once that is done, you should be able to use your model normally. To get a list of objects you'd do something like:
qry_desc_char_list = QryDescChar.objects.all()
To actually get the list into your template you might want to look at generic views, specifically the object_list view.
If your RDBMS lets you create writable views and the view you create has the exact structure than the table Django would create I guess that should work directly.
(This is an old question, but is an area that still trips people up and is still highly relevant to anyone using Django with a pre-existing, normalized schema.)
In your SELECT statement you will need to add a numeric "id" because Django expects one, even on an unmanaged model. You can use the row_number() window function to accomplish this if there isn't a guaranteed unique integer value on the row somewhere (and with views this is often the case).
In this case I'm using an ORDER BY clause with the window function, but you can do anything that's valid, and while you're at it you may as well use a clause that's useful to you in some way. Just make sure you do not try to use Django ORM dot references to relations because they look for the "id" column by default, and yours are fake.
Additionally I would consider renaming my output columns to something more meaningful if you're going to use it within an object. With those changes in place the query would look more like (of course, substitute your own terms for the "AS" clauses):
CREATE VIEW qry_desc_char as
SELECT
row_number() OVER (ORDER BY tbl_char.cid) AS id,
tbl_desc.iid_id AS iid_id,
tbl_desc.cid_id AS cid_id,
tbl_desc.cs AS a_better_name,
tbl_char.cid AS something_descriptive,
tbl_char.charname AS name
FROM tbl_desc,tbl_char
WHERE tbl_desc.cid_id = tbl_char.cid;
Once that is done, in Django your model could look like this:
class QryDescChar(models.Model):
iid_id = models.ForeignKey('WhateverIidIs', related_name='+',
db_column='iid_id', on_delete=models.DO_NOTHING)
cid_id = models.ForeignKey('WhateverCidIs', related_name='+',
db_column='cid_id', on_delete=models.DO_NOTHING)
a_better_name = models.CharField(max_length=10)
something_descriptive = models.IntegerField()
name = models.CharField(max_length=50)
class Meta:
managed = False
db_table = 'qry_desc_char'
You don't need the "_id" part on the end of the id column names, because you can declare the column name on the Django model with something more descriptive using the "db_column" argument as I did above (but here I only it to prevent Django from adding another "_id" to the end of cid_id and iid_id -- which added zero semantic value to your code). Also, note the "on_delete" argument. Django does its own thing when it comes to cascading deletes, and on an interesting data model you don't want this -- and when it comes to views you'll just get an error and an aborted transaction. Prior to Django 1.5 you have to patch it to make DO_NOTHING actually mean "do nothing" -- otherwise it will still try to (needlessly) query and collect all related objects before going through its delete cycle, and the query will fail, halting the entire operation.
Incidentally, I wrote an in-depth explanation of how to do this just the other day.
You are trying to fetch records from a view. This is not correct as a view does not map to a model, a table maps to a model.
You should use Django ORM to fetch QryDescChar objects. Please note that Django ORM will fetch them directly from the table. You can consult Django docs for extra() and select_related() methods which will allow you to fetch related data (data you want to get from the other table) in different ways.