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How i can realize multiple pagination(drf)

If I'll send get request like thisenter image description here, i need to have multiple pagination (LimitOffset and PageNumber).
models.py:
from django.db import models
class Products(models.Model):
title = models.CharField(max_length=255)
description = models.TextField(blank=True)
photo = models.ImageField(upload_to="photos/%Y/%m/%d/", null=True)
hashtag = models.CharField(max_length=255)
is_hit = models.BooleanField(default=False)
category = models.ForeignKey('Category', on_delete=models.PROTECT, null=True)
def __str__(self):
return self.title
class Category(models.Model):
name = models.CharField(max_length=255)
def __str__(self):
return self.name
views.py:
from rest_framework import generics
from rest_framework.pagination import PageNumberPagination
from .models import *
from .serializers import ProductsSerializer
class PaginationProducts(PageNumberPagination):
page_size = 2
page_size_query_param = 'page_size'
max_page_size = 2
class ProductsAPIList(generics.ListCreateAPIView):
queryset = Products.objects.all()
serializer_class = ProductsSerializer
pagination_class = PaginationProducts
serializers.py
from rest_framework import serializers
from .models import *
class ProductsSerializer(serializers.ModelSerializer):
class Meta:
model = Products
fields = "__all__"
def get_photo_url(self, obj):
request = self.context.get('request')
photo_url = obj.fingerprint.url
return request.build_absolute_uri(photo_url)
I need something that can help API client choose number of page and quantity of posts on that page. Think that in this case i need NumberPagePagination and LimitOffsetPagination.

I think you don't need to create the custom pagination class.
from rest_framework.pagination import LimitOffsetPagination
class ProductsAPIList(generics.ListCreateAPIView):
queryset = Products.objects.all()
serializer_class = ProductsSerializer
pagination_class = LimitOffsetPagination
The offset value corresponds to the page * size in the original pagination and the limit value corresponds to the size.

Django + Postgres + UUID

UPDATE:
I have got it to work!
What I did was:
Deleted the app
Went into Postgres Admin and deleted the Schema for that app
Created the app again (using a diff name)
Created the Models, URLs, Admin Registering, Views
So I've been trying to add a UUID field to my model, to use it in my URL but I keep getting this error
django.db.utils.ProgrammingError: column "id" is of type bigint but expression is of type uuid
LINE 1: ..._book" ("id", "title", "author", "price") VALUES ('f6b15400-...
^
HINT: You will need to rewrite or cast the expression.
Here is my models.py:
from django.db import models
import uuid
from django.urls import reverse
# Create your models here.
class Book(models.Model):
id = models.UUIDField(primary_key=True, unique=True, default=uuid.uuid4, editable=False)
title = models.CharField(max_length=200)
author = models.CharField(max_length=200)
price = models.DecimalField(max_digits=6, decimal_places=2)
def __str__(self):
return self.title
def get_absolute_url(self):
return reverse('book_detail', args=[str(self.id)])
Here is the views.py:
from django.shortcuts import render
from django.views.generic import ListView, DetailView
from .models import Book
# Create your views here.
class BookListView(ListView):
model = Book
template_name = 'books/book_list.html'
context_object_name = 'book_list'
class BookDetailView(DetailView):
model = Book
template_name = 'books/book_detail.html'
context_object_name = 'book'
Here is the urls.py:
from django.urls import path
from .views import BookListView, BookDetailView
urlpatterns = [
path('', BookListView.as_view(), name='book_list'),
path('<uuid:pk>/', BookDetailView.as_view(), name='book_detail'),
]

First add the uuid field as a normal field with a different name:
from django.db import models
import uuid
from django.urls import reverse
class Book(models.Model):
uuid = models.UUIDField(default=uuid.uuid4, unique=True)
title = models.CharField(max_length=200)
author = models.CharField(max_length=200)
price = models.DecimalField(max_digits=6, decimal_places=2)
Run makemigrations and migrate
Now make it primary key:
class Book(models.Model):
uuid = models.UUIDField(default=uuid.uuid4, primary_key=True)
title = models.CharField(max_length=200)
author = models.CharField(max_length=200)
price = models.DecimalField(max_digits=6, decimal_places=2)
This should generate a migration that removes the id field and makes uuid primary key (makemigrations and migrate again).
And finally:
class Book(models.Model):
id = models.UUIDField(primary_key=True, unique=True, default=uuid.uuid4, editable=False)
title = models.CharField(max_length=200)
author = models.CharField(max_length=200)
price = models.DecimalField(max_digits=6, decimal_places=2)
You may have to adjust the generated migration.

Django Forms created from Model with overfull Queryset Output

I am trying to create a ModelForm for my Model Class "Asset" in Django 3
from django.db import models
from django.contrib.auth.models import User
from django.forms import ModelForm
class Manufacturer(models.Model):
name = models.CharField(max_length=100)
class Asset(models.Model):
serial = models.CharField(max_length=200)
manufacturer = models.ManyToManyField(Manufacturer)
author = models.ForeignKey(User, on_delete=models.SET_NULL, null=True)
name = models.CharField(max_length=200)
I managed to create a Form via the following code
from django import forms
from .models import Manufacturer
class AssetForm(forms.Form):
serial = forms.CharField(max_length=150)
manufacturer = forms.ModelChoiceField(queryset=Manufacturer.objects.all().values('name'))
name = forms.CharField(max_length=200)
description = forms.TextInput()
status = forms.CharField(max_length=200)
category = forms.CharField(max_length=200)
The querySet results in a dropdown being filled out with either "{'name':'Apple'}" or "('Apple',)" depending on using values or values_list respectively. How can I just display the name itself?

Adding the following method to the model fixes the problem:
def __str__(self):
return self.name
This will return the name and only the name to the queryset.

django foreignkey not working

I am still not satisfied with django foreignkey relationship.
below are two classess .
from django.db import models
from django.utils.text import slugify
from django_countries.fields import CountryField
from django.conf import settings
# Create your models here.
class PhotoGallery(models.Model):
title = models.CharField(max_length=255,unique=True)
picture_choices =
models.CharField(max_length=255,choices=PICTURE_CHOICES,default=NTR)
description = models.TextField(default='')
user = models.ForeignKey(settings.AUTH_USER_MODEL,
on_delete=models.CASCADE,related_name='image_uploader')
date_posted = models.DateTimeField(auto_now=True)
views = models.PositiveIntegerField()
likes = models.PositiveIntegerField()
country = CountryField(blank_label='(select country)')
slug = models.SlugField(allow_unicode=True,unique=True)
def __str__(self):
return self.title
from django.contrib import admin
from gallery.models import PhotoGallery
class PhotoGalleryAdmin(admin.ModelAdmin):
prepopulated_fields = {'slug': ('title',)}
And below is class Imagesmodel that has a foreignkey .Calling the PhotoGallery class from class Imagesmodel is not a problem .MY big problem is doing the reverse
class ImagesModel(models.Model):
gala_obj =
models.ForeignKey(PhotoGallery,
related_name='picture_gallery',on_delete=models.CASCADE)
post_image = models.ImageField(upload_to='gallery_pics',
verbose_name='Image')
def __str__(self):
return self.gala_obj.title
and below is my views.py file
`def detailViewPage(request,slug):
#here i wanted to get one image with the given slug
from `ImagesModel`
a = get_object_or_404(PhotoGallery, slug=slug)
selected_pic = a.picture_gallery.all()
#here i wanted to get all images excluding one
with the given slug
object_2 = ImagesModel.objects.exclude(id=selected_pic.id)
.order_by('-id')[:15]
return render(request,'gallery/photogallery_detail.html',{
'selected_pic':selected_pic,
'object_2':object_2,
})`
below is the error i get

The question is not very clear, but I assume you want to get all pictures associated with an object A. In your B model, give your foreignkey a related name:
class B(models.Model):
a = models.ForeignKey(A,related_name='post_title', related_name='pictures')
image = models.ImageField(upload_to='gallery_pics',
verbose_name='Image')
In your views.py:
a = get_object_or_404(A, slug=slug)
b = a.pictures.all()
I don't know why you named your foreignkey 'title', but the above code will get all the pictures associated with the object A.
The following will work also:
a = get_object_or_404(A, slug=slug)
pictures = a.b_set.all()
If you want to get the slug from an object B, then you can do the following:
b = get_object_or_404(B, id=1)
slug = b.a.slug

Django: ManyToMany association name not defined

this is my models.py
from django.db import models
from django.urls import reverse
from django.contrib.auth.models import User
from userprofiles.models import UserProfile
# Create your models here.
import os
class Category (models.Model):
name = models.CharField(max_length=100)
def __str__(self):
return self.name
class Image(models.Model):
owner = models.ForeignKey(UserProfile, on_delete=models.CASCADE)
category = models.ForeignKey(Category, on_delete=models.CASCADE)
in_gallery = models.ManyToManyField(Gallery)
title = models.CharField(max_length=100)
def __str__(self):
return self.title
class Gallery(models.Model):
name = models.CharField(max_length=100)
have_images = models.ManyToManyField(Image)
def __str__(self):
return self.name
class ImageInGallery (models.Model):
image = models.ForeignKey(Image)
gallery = models.ForeignKey(Gallery)
line 22, in Image in_gallery = models.ManyToManyField(Gallery)
NameError: name 'Gallery' is not defined
I want to have manytomany relation for Image and Gallery and class ImageInGallery to store the Image~Gallery relation sets.
also, I want images belonging to a certain gallery accessible from a gallery and vice versa, so I put both models.ManyToManyField(Gallery) and models.ManyToManyField(Image) in both classes
whats the reason of the error:
line 22, in Image in_gallery = models.ManyToManyField(Gallery)
NameError: name 'Gallery' is not defined

Place the Gallery model above Image
Ex:
from django.db import models
from django.urls import reverse
from django.contrib.auth.models import User
from userprofiles.models import UserProfile
# Create your models here.
import os
class Category (models.Model):
name = models.CharField(max_length=100)
def __str__(self):
return self.name
class Gallery(models.Model):
name = models.CharField(max_length=100)
have_images = models.ManyToManyField(Image)
def __str__(self):
return self.name
class Image(models.Model):
owner = models.ForeignKey(UserProfile, on_delete=models.CASCADE)
category = models.ForeignKey(Category, on_delete=models.CASCADE)
in_gallery = models.ManyToManyField(Gallery)
title = models.CharField(max_length=100)
def __str__(self):
return self.title
class ImageInGallery (models.Model):
image = models.ForeignKey(Image)
gallery = models.ForeignKey(Gallery)

You have duplicated relationship between Image and Gallery.
class Image(models.Model):
in_gallery = models.ManyToManyField(Gallery)
class Gallery(models.Model):
have_images = models.ManyToManyField(Image)
You only need to define one of those fields and then you can use it like the docs describe it.
If you remove the field Gallery.have_images then you can have:
# list all galleries a image is part of
image_instance.in_gallery.all()
# list all images from a gallery
gallery_instance.image_set.all()

Just keep your structure the way you want, and put the model name inside quote
Like:
have_images = models.ManyToManyField("Image")
if the models is in another app, use the name of the app
Ex.
have_images = models.ManyToManyField("app_name.Image")