Develop Reference

display objects from inline model in DetailView

I have 2 models Product and Resource. Resource has to be a TabularInline model in my admin panel. I am struggling with filtering resource titles that are related only to this product. Since it is a ForeignKey I should use select_related but I am not sure how to use it in my case. For now, the loop in my HTML file gives me all sales files (from all products).
models.py
class Product(models.Model):
id = models.AutoField(primary_key=True)
title = models.CharField('title', max_length=400, default='')
slug = models.SlugField(unique=True, blank=True, max_length=600)
class Resource(models.Model):
id = models.AutoField(primary_key=True)
type = models.CharField(max_length=32, choices=RESOURCE_TYPE, default='sales')
title = models.CharField(max_length=400, blank=False, null=False, default='')
related_files = models.FileField(upload_to='recources/', null=True, blank=True)
publish = models.DateTimeField('Published on', default=timezone.now)
resources = models.ForeignKey(Product, default='', on_delete=models.PROTECT, null=True, related_name='resources')
admin.py
class Resourceinline(admin.TabularInline):
model = Resource
class ProductAdmin(ImportExportModelAdmin):
inlines = [
Resourceinline,
]
resource_class = ProductResource
admin.site.register(Product, ProductAdmin)
views.py
class ProductDetailView(DetailView):
template_name = 'ProductDetailView.html'
model = Product
def get_context_data(self, **kwargs):
context = super(ProductDetailView, self).get_context_data(**kwargs)
resources_sales = Resource.objects.select_related('resources').filter(resources_id =1, type='sales') # not sure what to put here
context['resources_sales'] = resources_sales
return context
ProductDetailView.html
{% for resource in resources_sales.all %}
<p>{{resource.title}}</p>
{% endfor %}
Question
Where am I making the mistake and how can I display resource objects that are related to type=sales and are related only to this product in DetailView.
Edit
I realized that there is a column named resources_id that is connecting both models. Now I am struggling to filter it by id of current DetailView. I put resources_id=1 in my views.py but it must relate to DetailView that user is currently looking at. I tied to put resources_id=self.kwargs['id'] but it gives me KeyError at /product/test-product/ 'id' How can I do that?

since you are using generic DetailView you can refer to the current object with self.get_object(). actually that return the single object that view display. however you can use instateself.object too.
so you can filter the Product related Resources using Resource.objects.filter(resources=self.get_object(), type='sales')
you can read more Single object mixins

How to edit a property in Django Admin?

I have a model with an attribute that is connected to another model as follow:
class Book(models.Model):
synced = models.OneToOneField('SyncedBook'
related_name='internal',
on_delete=models.CASCADE)
# some more attributes here...
#property
def book_address(self)
return self.synced.book_address
However, the book_address is a also a FK in the SyncedBook table as follow:
book_address = models.ForeignKey('Address', db_index=True, null=True, blank=True,
related_name='address_book', on_delete=models.PROTECT)
I don't know and understand how to be able to edit the book_address through the Django admin page in class BookingAdmin(admin.ModelAdmin), even though I have read over the documentation. At first I have the attribute as readonly, but now I want to be able to edit it and save the new address from the Address table. Is there a way to make it happen through the class BookingAdmin(admin.ModelAdmin) and how? Any example and solution would be appreciate

Model properties are typically used for presenting logically defined data for a particular model instance and not necessarily storing data on the model instance itself.
An example of when to use a model property is as follows:
# Defines a product instance
class Product(model.Models):
name = models.CharField(max_length=100)
description = models.TextField()
active = models.BooleanField(default=True)
cost = models.DecimalField(max_digits=5, decimal_places=2)
price = models.DecimalField(max_digits=5, decimal_places=2)
# calculate profits on product
#property
def profit(self)
p = self.price - self.cost
return p
In your case, you are trying to actually be able to modify data against a related model instance within the django admin. To me this sounds like more specifically an Inline (click here for documentation)
So in your case, you would need to create something like the following to your admin.py file:
class SyncedBookInline(admin.TabularInline):
model = BookInline
#admin.Register(Book)
class BookAdmin(admin.ModelAdmin):
# all your model admin settings
inlines = [SyncedBookInline]
Additional Info:
The Inline solution should still work for you. Please see the working code listed below:
models.py:
from django.db import models
class Hero(models.Model):
name = models.CharField(max_length=50)
class HeroAcquaintance(models.Model):
name = models.CharField(max_length=50)
hero = models.OneToOneField(Hero, on_delete=models.CASCADE)
admin.py:
from django.contrib import admin
from .models import *
class HeroAcquaintanceInline(admin.TabularInline):
model = HeroAcquaintance
#admin.register(Hero)
class HeroAdmin(admin.ModelAdmin):
list_display = (
'name',
)
inlines = [HeroAcquaintanceInline]
#admin.register(HeroAcquaintance)
class HeroAcquaintanceAdmin(admin.ModelAdmin):
list_display = (
'name',
)
Screenshot:

Having admin inline for model without direct relation or foreign key

So, I have this these model:
class Media(models.Model):
title = models.CharField(max_length=50, blank=True)
slug = models.SlugField(max_length=70, blank=True, editable=False)
class Meta:
abstract = True
class Photo(Media):
source = models.ImageField(upload_to='uploads/gallery/photo')
class Video(Media):
source = models.FileField(upload_to='uploads/gallery/photo')
class Item(models.Model):
content_type = models.ForeignKey(
ContentType,
limit_choices_to={'model__in': ['photo', ]},
on_delete=models.CASCADE,
)
object_id = models.PositiveIntegerField()
content_object = GenericForeignKey('content_type', 'object_id')
class Album(Media):
items = models.ManyToManyField(Item)
How can I have an album admin with photo and video as inline so i can upload photo and video when i created an album ?
When I tried making inline for photo and hook it to album admin, I get error "photo has no foreignkey to album", pretty obvious, from there I think there should be a way to link foreignkey needed by album admin with content object from model item.
Note: I specifically doesn't want an item admin. An item are created at model post save signals.

I do not think this works out of the box. But you can use libraries like eg django-smart-selects-generics. Depending on your django version you might need to update some files there.
Installation works with:
pip install django-smart-selects-generic
You also need to have django-smart-selects installed.
Then add both apps in the settings.
INSTALLED_APPS = (
...
'smart_selects',
'smart_generic'
)
And then in your admin.py you can do:
from smart_generic.form_fields import GenericChainedModelChoiceField
from django.forms.models import ModelForm
class TForm(ModelForm):
object_id = GenericChainedModelChoiceField('content_type','content_type',label=u'Content object',queryset=TargetRelation.objects.all())
class Meta:
model = TargetRelation
fields = ['target','content_type']
class TRAdmin(admin.ModelAdmin):
form = TForm
class TRInline(admin.TabularInline):
model = TargetRelation
form = TForm
class PlanetAdmin(admin.ModelAdmin):
inlines=[TRInline,]
Depending on your django version you might need to replace in widgets.py:
Media = ChainedSelect.Media
by
.media = ChainedSelect.Media
And in smart_select views.py add:
import json as simplejson
and replace the return statement by:
return HttpResponse(json, content_type='application/json')

How can I create Django ModelForm for an abstract Model?

I have an abstract model that all my other models inherit from, it looks like this.
class SupremeModel(models.Model):
creator = models.ForeignKey(User, related_name="%(class)s_creator")
created = models.DateTimeField(null=True, blank=True)
deleted = models.BooleanField(default=False)
modified = models.DateTimeField(null=True,blank=True)
class Meta:
abstract = True
I then have a bunch of other models that inherit from this model, with something along these lines...
class ExampleModel(SupremeModel):
name = models.TextField(null=False, blank=False)
description = models.TextField(null=False, blank=False)
class AnotherModel(SupremeModel):
title = models.TextField(null=False, blank=False)
location = models.TextField(null=False, blank=False)
I want to create a Django model form for nearly all of my custom models that look similar to ExampleModel, but I always want the fields in SupremeModel to be excluded in the form...
How can I create a ModelForm that can be used to inherit the exclude parameters that will hide creator,created,deleted, and modified but show all of the other fields (in this case name and description or title and location).

you may try this
class ExcludedModelForm(ModelForm):
class Meta:
exclude = ['creator', 'created', 'deleted', 'modified']
class ExampleModelForm(ExcludedModelForm):
class Meta(ExcludedModelForm.Meta):
model = ExampleModel

How to display choices in my form using django forms?

I have a model that looks like this in my application:
class Member(models.Model):
name = models.CharField(max_length=200)
telephone_number = models.CharField(max_length=200)
email_address = models.CharField(max_length=200)
membership_type = models.CharField(max_length=1, choices=MEMBERSHIP_TYPES)
membership_number = models.CharField(max_length=200, primary_key=True)
def __unicode__(self):
return self.name
Reading up on forms for my model i see that i can replace the models stuff with forms and place it in my forms.py. However when for membership type i don't know how it is used in the forms class.

Hmmm... You might want to read on ModelForms.

Using the multiple choice field in the forms class.