I would like to ask how I could list all objects of logged current user via class based view in django.
I have two apps in the project. One is called users and the other one is badminton.
users/models.py
from django.db import models
from django.contrib.auth.models import User
class Profile(models.Model):
user = models.OneToOneField(User, null=True, on_delete=models.CASCADE)
name = models.CharField(max_length=200, null=True)
badminton/models.py
from django.db import models
from users import models as users_models
class Player(models.Model):
name = models.OneToOneField(users_models.Profile ,null=True, on_delete=models.CASCADE)
matches_played = models.IntegerField(default=0, blank=True, null=True)
class Match(models.Model):
player_home = models.OneToOneField(Player, null=True, on_delete= models.SET_NULL, related_name='player_home')
player_away = models.OneToOneField(Player, null=True, on_delete= models.SET_NULL, related_name='player_away')
How I can access all matches of logged user via queryset? Thank you for your help!
You can list Matches for which the request.user has as Player object player_home and/or player_away for that match with:
from django.db.models import Q
Match.objects.filter(
Q(player_home__name__user=request.user) |
Q(player_away__name__user=request.user)
)
Note: A foreign key refers to an object, not to the string representation of that
object. Therefore name is not really a good name to refer to a Profile object. You
might want to consider renaming it to profile.
Related
I am trying to create an announcement website (All) that can be visible to others (the Users, for which I added an Account). For this I wanted to modify a little the user profile to add fields like telephone, email address...
So I modified admin.py:
from django.contrib import admin
from .models import Todo, Account
from django.contrib.auth.models import User
class AccountInline(admin.StackedInline):
model = Account
can_delete = False
verbose_name_plural = 'Accounts'
class TodoAdmin(admin.ModelAdmin):
readonly_fields = ('created',)
inlines = (AccountInline, )
admin.site.unregister(User)
admin.site.register(Todo, TodoAdmin)
But got back:
<class 'todo.admin.AccountInline'>: (admin.E202) 'todo.Account' has no ForeignKey to 'todo.Todo'.
So I added a ForeignKey to Todo with account = models.ForeignKey(Account, on_delete=models.CASCADE):
from django.db import models
from django.contrib.auth.models import User
class Account(models.Model):
user = models.OneToOneField(User, on_delete=models.CASCADE)
email = models.CharField(max_length=100)
firstname = models.CharField(max_length=30)
lastname = models.CharField(max_length=50)
company = models.CharField(max_length=5)
def __str__(self):
return self.user.username
class Todo(models.Model):
title = models.CharField(max_length=100)
datetime = models.DateTimeField()
memo = models.TextField(blank=True)
created = models.DateTimeField(auto_now_add=True)
datecompleted = models.DateTimeField(null=True, blank=True)
important = models.BooleanField(default=False)
user = models.ForeignKey(User, on_delete=models.CASCADE)
account = models.ForeignKey(Account, on_delete=models.CASCADE)
def __str__(self):
return self.title
But I still have the error, and I don't have any Users in the admin panel anymore
You accidentally wrote unregister for Users in your admin.py file. It should be admin.site.register(User)
You misinterpretted the error: the error states that you don't have a foreign key in your Account model to Todo.
This means your inline admin code isn't correct as it's expecting the other way around.
I want to assign a User to a Model in django, I created a custom User model and sign-up/sign-in Forms but now I want to Assign a User model to another model named Customer whenever a new user is Created Here he the Customer model
class Customer(models.Model):
id = models.AutoField(primary_key=True)
User = models.OneToOneField(
Account, on_delete=models.CASCADE, null=True, blank=True)
name = models.CharField(max_length=200, null=True)
email = models.CharField(max_length=200, null=True)
phone = models.CharField(max_length=200, default='0', null=True, blank=True)
address = models.CharField(
max_length=200, default=' ', null=True, blank=True)
city = models.CharField(max_length=200, default=' ', null=True, blank=True)
def __str__(self):
if self.name == None:
return "ERROR-CUSTOMER NAME IS NULL"
return self.name
Note: I can assign the User manually in the Database and It lists All the Users but I want it to do it itself when a new user is created
I think it would be better to extend the User model, and add more fields rather than creating a new model (which has a User onetoonefiled in it).
Something like this:
from django.db import models
from django.contrib.auth.models import User
class Profile(models.Model):
user = models.OneToOneField(User, on_delete=models.CASCADE)
bio = models.TextField(max_length=500, blank=True)
location = models.CharField(max_length=30, blank=True)
birth_date = models.DateField(null=True, blank=True)
This is the kind of approach I use in my projects.
Here you have the default User model fields:
User model default fields
You don't need to add these in your Profile class.
I based this on this article: How to extend User Django Model
don't forget to add to the admin.py:
from django.contrib import admin
from .models import Profile
# Register your models here.
admin.site.register(Profile)
to see the Profiles in the admin page
Got it fixed by setting the user in the Customer model when a user is created
Customer.objects.create(user=request.user, name=username, email=email, phone=phone)
I need advice on a multiple user type.
Scenario:
A user can be an organization, where in this organization they can place adverts on the website. The owner of this organization(user) can edit/delete users and adverts of his own organization(group). In this organization user type there are users that also can log in and they can only see the adverts placed by them, but the owner of this group must see all adverts of his own and of his users. Think like an estate listing where an organization has multiple locations/users that can place adverts, and has to be managed by a admin user of this organization.
What type or model is the best/cleanest for implementing this in a good way? Do I need the Django's User and Group model?
One solution would be to have the "User Profiles" structure.
So you use the standard Django User Model and you attach to it several OneToOne relationships depending on the number of profile types you'll have. This has the advantage of allowing users to cover more than one role at the same time.
For example:
from django.contrib.auth.models import User
class Organization(models.Model):
user = models.OneToOneField(User, on_delete=models.CASCADE, related_name="organization")
name = models.CharField(max_length=50, blank=True, null=True)
class Supervisor(models.Model):
user = models.OneToOneField(User, on_delete=models.CASCADE, related_name="supervisor")
name = models.CharField(max_length=50, blank=True, null=True)
organization = models.ForeignKey(Organization, on_delete=models.CASCADE, related_name="supervisors")
class CustomUser(models.Model):
user = models.OneToOneField(User, on_delete=models.CASCADE, related_name="custom_user")
name = models.CharField(max_length=50, blank=True, null=True)
supervisor = models.ForeignKey(Supervisor, on_delete=models.CASCADE, related_name="custom_users", blank=True, null=True)
And then when you go and create the models for the ads to be displayed on the website you can use the built-in PermissionRequiredMixin.
In order to do that you have to start by adding "permissions" in the ad model Meta class:
class Ad(models.Model):
# fields
class Meta:
permissions = [
('can_edit_ads', 'org_representative')
]
Then on your view you have to extend the PermissionRequiredMixin, example:
class EditAd(UpdateView, PermissionRequiredMixin):
model = Ad
template_name = "ad123.html"
permission_required = "ad.can_edit_ads"
A quick way to test it is by going in the user table on the admin panel, open a user detail page where you can see all the permissions, and there alongside the others you'll find your custom one as well.
From there you can easily assign the new permission to the specific user.
I have two models:
from django.db import models
from django.contrib.auth.models import User
# Create your models here.
class Category(models.Model):
user = models.ForeignKey(User, on_delete=models.CASCADE, related_name="categories")
name = models.CharField(max_length=30, unique=True, primary_key=True)
class Todo(models.Model):
user = models.ForeignKey(User, on_delete=models.CASCADE, related_name='todos')
# TODO: Add confirmation before deleting category
category = models.ForeignKey(Category, on_delete=models.CASCADE,
related_name="todos_in_category", null=True)
item = models.CharField(max_length=50)
added = models.DateTimeField(auto_now_add=True)
completed = models.BooleanField(default=False)
Previously, Category's PK was the default id, however, I changed it to the name field. When I ran the migrations, i received the operational error. Thinking that it was perhaps due to a conflict between the existing id fields and the new primary key, I cleared the data in the database but with no success. Any ideas as to what could be the issue here? Thanks!
My admin page is working fine except when logged in it is not showing any user models. It is hindering my work as I cannot manage users.
I have made custom models as shown below.
Database is MySQL.
models.py
class User(AbstractUser):
is_customer = models.BooleanField(default=False)
is_restaurant = models.BooleanField(default=False)
first_name = models.CharField(max_length=100)
last_name = models.CharField(max_length=100)
class Customer(models.Model):
user = models.OneToOneField(User, on_delete=models.CASCADE, primary_key=True)
food_pref = models.CharField(max_length=10, default='veg')
class Restaurant(models.Model):
user = models.OneToOneField(User, on_delete=models.CASCADE, primary_key=True)
restaurant_name = models.CharField(max_length=100, blank=False)
Regisrar your models inadmin.py file.
from . models import Model_Name
Then you can register your models in two ways:
I) admin.site.register(Model_Name)
II)
#admin.register(Model_Name)
Class Xyz(admin.ModelAdmin):
pass
Second method gives you more flexibility like list_display, list_filter, date_hierarchy, etc. for customising your Admin section/site.
You can look more about customising admin site at https://docs.djangoproject.com/en/3.0/ref/contrib/admin/#module-django.contrib.admin
Have You registered them in admin.py?
from .models import ModelName
admin.site.register(ModelName)