Develop Reference

Access several tables with only one query

I have the same schema in my django application:
class SomeModel(models.Model):
value = models.CharField(max_length=30)
class AbstractModel(models.Model):
someModel = models.ForeignKey(SomeModel)
class Meta:
abstract = True
class A(AbstractModel):
anotherValue = models.CharField(max_length=5)
class B(AbstractModel):
anotherValue = models.CharField(max_length=5)
class C(AbstractModel):
anotherValue = models.CharField(max_length=5)
class D(AbstractModel):
anotherValue = models.CharField(max_length=5)
class E(AbstractModel):
anotherValue = models.CharField(max_length=5)
With this layout, I need the most efficient way to query all objects from models A, B, C, D and E with a given id of SomeModel. I know that I cannot execute a query in an abstract model, so right now, what I do is query each model separately like this:
A.objects.filter(someModel__id=id)
B.objects.filter(someModel__id=id)
C.objects.filter(someModel__id=id)
D.objects.filter(someModel__id=id)
E.objects.filter(someModel__id=id)
Obviously this approach is quite slow, because I need to make 5 different queries each time I want to know all those objects. So my question is, is there a way to optimize this kind of query?
UPDATE:
I have tried the union method like this:
qs1 = A.objects.filter(**filters) # hits DB
qs2 = B.objects.filter(**filters) # hits DB
qs3 = C.objects.filter(**filters) # hits DB
qs4 = D.objects.filter(**filters) # hits DB
qs5 = E.objects.filter(**filters) # hits DB
qs1.union(qs2, qs3, qs4, qs5) # hits DB
That's actually 6 hits to the database!! I woulk like only one!
I have checked this printing the number of queries made:
from django.conf import settings
settings.DEBUG = True
from django.db import connection
print(len(connection.queries))

You may use union method, but what you want to do? If you want to call five objects by one pk and you want to be sure that they have strict relation between each other you may use OneToOne relationship.
So in the first case you just need to make a query, in the second case you must make new migration and maybe you will need to rebuild your tables.

How to update queryset value in django?

I have written a python script in my project. I want to update the value of a field.
Here are my modes
class News_Channel(models.Model):
name = models.TextField(blank=False)
info = models.TextField(blank=False)
image = models.FileField()
website = models.TextField()
total_star = models.PositiveIntegerField(default=0)
total_user = models.IntegerField()
class Meta:
ordering = ["-id"]
def __str__(self):
return self.name
class Count(models.Model):
userId = models.ForeignKey(User, on_delete=models.CASCADE)
channelId = models.ForeignKey(News_Channel, on_delete=models.CASCADE)
rate = models.PositiveIntegerField(default=0)
def __str__(self):
return self.channelId.name
class Meta:
ordering = ["-id"]
This is my python script:
from feed.models import Count, News_Channel
def run():
for i in range(1, 11):
news_channel = Count.objects.filter(channelId=i)
total_rate = 0
for rate in news_channel:
total_rate += rate.rate
print(total_rate)
object = News_Channel.objects.filter(id=i)
print(total_rate)
print("before",object[0].total_star,total_rate)
object[0].total_star = total_rate
print("after", object[0].total_star)
object.update()
After counting the total_rate from the Count table I want to update the total star value in News_Channel table. I am failing to do so and get the data before the update and after the update as zero. Although total_rate has value.

The problem
The reason why this fails is because here object is a QuerySet of News_Channels, yeah that QuerySet might contain exactly one News_Channel, but that is irrelevant.
If you then use object[0] you make a query to the database to fetch the first element and deserialize it into a News_Channel object. Then you set the total_star of that object, but you never save that object. You only call .update() on the entire queryset, resulting in another independent query.
You can fix this with:
objects = News_Channel.objects.filter(id=i)
object = objects[0]
object.total_star = total_rate
object.save()
Or given you do not need any validation, you can boost performance with:
News_Channel.objects.filter(id=i).update(total_star=total_rate)
Updating all News_Channels
If you want to update all News_Channels, you actually better use a Subquery here:
from django.db.models import OuterRef, Sum, Subquery
subq = Subquery(
Count.objects.filter(
channelId=OuterRef('id')
).annotate(
total_rate=Sum('rate')
).order_by('channelId').values('total_rate')[:1]
)
News_Channel.objects.update(total_star=subq)

The reason is that object in your case is a queryset, and after you attempt to update object[0], you don't store the results in the db, and don't refresh the queryset. To get it to work you should pass the field you want to update into the update method.
So, try this:
def run():
for i in range(1, 11):
news_channel = Count.objects.filter(channelId=i)
total_rate = 0
for rate in news_channel:
total_rate += rate.rate
print(total_rate)
object = News_Channel.objects.filter(id=i)
print(total_rate)
print("before",object[0].total_star,total_rate)
object.update(total_star=total_rate)
print("after", object[0].total_star)

News_Channel.total_star can be calculated by using aggregation
news_channel_obj.count_set.aggregate(total_star=Sum('rate'))['total_star']
You can then either use this in your script:
object.total_star = object.count_set.aggregate(total_star=Sum('rate'))['total_star']
Or if you do not need to cache this value because performance is not an issue, you can remove the total_star field and add it as a property on the News_Channel model
#property
def total_star(self):
return self.count_set.aggregate(total_star=Sum('rate'))['total_star']

NameError in Django simple search

I have a simple search in my Django project. I want to search through documents using their type and part of factory info in addition to search by name.
Here is my models.py:
class Docs(models.Model):
Date = models.DateField(default=date.today)
Name = models.CharField(max_length=50)
Type = models.ForeignKey(DocTypes)
Part = models.ForeignKey(Parts)
Link = models.FileField(upload_to='Docs/%Y/%m/%d')
class Parts(models.Model):
Name = models.CharField(max_length=50)
def __str__(self):
return str(self.Name)
class DocTypes(models.Model):
Type = models.CharField(max_length=50)
def __str__(self):
return str(self.Type)
My forms.py:
class DocsSearchForm(ModelForm):
class Meta:
model = Docs
fields = [ 'Name', 'Type', 'Part']
And this is part of my views.py, if no search was done then all documents are given
def showdocs(request):
if request.method == 'POST':
form = DocsSearchForm(request.POST)
documents = Docs.objects.filter(Name__contains=request.POST['Name']|
Type==request.POST['Type']|
Part==request.POST['Part'])
else:
form = DocsSearchForm()
documents = Docs.objects.all()
return render(
request,
'showdocs.html',
{'documents': documents, 'form':form}
So, the problem is the following: if I try to use a search then I have
NameError at /showdocs
name 'Type' is not defined.
POST values are:Part '1', Name 'Example', Type '1'.
If I delete
Type==request.POST['Type']|
Part==request.POST['Part']
then search by name works well. So I have a guess that problem is about searching by foreign key values, but have no ideas more. Will appreciate any help.

Try replacing the line with this
Docs.objects.filter(Name__contains=request.POST['Name'],
Type=request.POST['Type'],
Part=request.POST['Part']
)
It seems you have misunderstood the syntax. I don't know why you are trying to use | operator here.

That's not how Django filters work. You can't | them because they are not actually expressions, just keyword arguments. In this case, correct syntax would be:
Docs.objects.filter(
Name__contains=request.POST['Name'],
Type_Type=request.POST['Type'],
Part_Name=request.POST['Part'],
)`

Django - using Foreign key mode attribute

I have two Django models and connected via Foreignkey element and in the second model I need to use the firs model's attribute - example (pseudocode):
Class Category(models.Model):
c_attribute = "Blue"
Class Object(models.Model):
o_category = models.ForeignKey(Category)
o_color = o_category.c_attribute
The key here is the last line - I got error saying that ForeignKey object has no attribute c_attribute.
Thanks

Because o_category is a Key to a Category, not a Category itself!
you can check by type(o_category) to check is not Category!
so you have access to related Cateogry of a Object in other parts of application when connected to database.for example in shell you can write:
c = Category()
c.save()
o = Object(o_category = c, ...) #create Object with desired params
... #some changes to o
o.save()
o.o_category.c_attribute #this will work! :)

You can use to_field='', but that might give you an error as well.
Class Object(models.Model):
o_category = models.ForeignKey(Category)
o_color = models.ForeignKey(Category, to_field="c_attribute")
The best thing is do create a function in your Object model, that would get you the categories c_attribute like so:
def get_c_attribute(self):
return self.o_category.c_attribute

Django ManyToManyField Error when saving in admin?

What is wrong with my code?
class Group(ImageModel):
title = models.CharField(verbose_name = "Title", max_length=7)
photos = models.ManyToManyField('Photo', related_name='+',
verbose_name=_('Photo'),
null=True, blank=True)
.....
pid = Photo.objects.get(image = str_path)
gid= Group.objects.get(id = self.id)
self.save_photos(gid, pid)
....
def save_photos(self, gid, pid):
group_photo = GroupPhotos(groupupload=gid.id,
photo=pid.id
)
group_photo.save()
and my GroupPhotos models is:
class GroupPhotos(models.Model):
groupupload = models.ForeignKey('Group')
photo = models.ForeignKey('Photo')
class Meta:
db_table = u'group_photos'
when i want to save it from admin panel i am getting value error sth like this:
Cannot assign "38": "GroupPhotos.groupupload" must be a "Group" instance.
with group_photo = GroupPhotos(groupupload=gid, photo=pid) defination it is working but there is no any changes in GroupPhotos table(group_photos). printing this print pid.id,' >>> ',gid.id i am getting true relation...
UPDATE:
I have been working since morning, but no progress... i have also tried this but nothing changed:
pid = Photo.objects.get(image = str_path)
ger = Group.objects.get(id = self.id)
ger.title = self.title
ger.save()
ger.photos.add(pid)

The error is here:
group_photo = GroupPhotos(groupupload=gid.id, photo=pid.id)
The arguments to groupupload and photo should be instances of Group and Photo respectively. Try the following:
group_photo = GroupPhotos(groupupload=gid, photo=pid)
In other words, when creating an object you need to pass arguments of the expected type and not an integer (which may be the primary key key of the desired object but it also might not, which is why you need to pass an object of the correct type).

i have solved my problem with adding through option to my manytomanyfield:
photos = models.ManyToManyField('Photo', related_name='+',
verbose_name=_('Photo'),
null=True, blank=True, through=GroupPhotos)
some info about ManyToManyField.through here:
Django will automatically generate a table to manage many-to-many
relationships. However, if you want to manually specify the
intermediary table, you can use the through option to specify the
Django model that represents the intermediate table that you want to
use.
The most common use for this option is when you want to associate extra data with a many-to-many relationship.