what field should I use for password? - python

I am using python version 3.5 and django version 1.11.
I wanted to know what field should I use for pasword
models.py
from django.db import models
from django.db.models.signals import post_save
from django.contrib.auth.models import User
class Employee(models.Model):
name = models.CharField(max_length=50,null=True)
username = models.CharField(max_length=100,null=True)
password = models.CharField(max_length=128)
repeat_password = models.CharField(max_length=128)
email = models.EmailField(max_length=70,null=True)
phone_number = models.IntegerField(default=0)
permanent_address = models.CharField(max_length=100,null=True)
alternative_address = models.CharField(max_length=100,null=True)
salary = models.IntegerField(default=0)
join_date = models.DateField(null=True)
paid_salary_date = models.DateField(null=True)
designation = models.CharField(max_length=50,null=True)
def __str__(self):
return self.name

Don't reinvent the wheel, extend the Django User model instead, and add those extra fields to your Employee class. You can equally override or define new methods.
See Extending the existing User model. Infact, the example provided in the docs is an Employee class.
from django.contrib.auth.models import User
class Employee(models.Model):
user = models.OneToOneField(User, on_delete=models.CASCADE)
department = models.CharField(max_length=100)

Related

How to connect 2 many to many fields in Django

I have 2 many to many fields in models and i want to connect them to each other i mean if i connect user in Admin Model with Counter Party i cant see that in Counter Party admin
How can i do that?
When im trying to do that it shows only in 1 model
models.py
class CustomUser(models.Model):
user = models.OneToOneField(User, on_delete=models.CASCADE, verbose_name='Пользователь')
user_counter = models.ManyToManyField('CounterParty', blank=True, verbose_name='Контрагенты пользователя')
def __str__(self):
return f'{self.user}'
class CounterParty(models.Model):
GUID = models.UUIDField(default=uuid.uuid4, editable=True, unique=True)
name = models.CharField(max_length=150, verbose_name='Наименование')
customer = models.BooleanField(default=False, verbose_name='Заказчик')
contractor = models.BooleanField(default=False, verbose_name='Подрядчик')
counter_user = models.ManyToManyField(User, blank=True, related_name='counter_user',
verbose_name='Пользователи контрагента')
class Meta:
verbose_name = 'Контрагент'
verbose_name_plural = 'Контрагенты'
def __str__(self):
return
admin.py
from django.contrib import admin
from .models import CustomUser, CounterParty, ObjectList, SectionList
from authentication.models import User
from authentication.admin import UserAdmin
class CustomUserInLine(admin.StackedInline):
model = CustomUser
can_delete = False
verbose_name_plural = 'Пользователи'
class CustomUserAdmin(UserAdmin):
inlines = (CustomUserInLine,)
#admin.register(CounterParty)
class CounterPartyAdmin(admin.ModelAdmin):
pass
admin.site.unregister(User)
admin.site.register(User, CustomUserAdmin)
user admin
counter party admin
You would not want to have these kinds of references with ManyToMany. Ideally you would have a one sided reference.
You can do an inline in your admin like this:
class CustomUserInLine(admin.StackedInline):
model = "CustomUser.user_counter.through"
Here are the docs for inline M2M in the admin: Django docs

UserProfile not creating in custom model using User model as OneToOneField(User)

I am creating the super user through admin panel or command line, but it is not showing in the custom model in the database.
models.py
from django.db import models
from django.contrib.auth.models import User
# Create your models here.
class Profile(models.Model):
user = models.OneToOneField(User, on_delete=models.CASCADE)
description = models.TextField(null=True)
profile_image = models.ImageField(null=True, blank=True)
def __str__(self):
return str(self.user)
class Following(models.Model):
user = models.ForeignKey(User, on_delete=models.CASCADE)
following = models.CharField(max_length=30)
class Follower(models.Model):
user = models.ForeignKey(User, on_delete=models.CASCADE)
follower = models.CharField(max_length=30)
Using OneToOnefield should create the user in the Profile model automatically but it is not happening. I am not sure what is wrong because in the previous project it was workiing fine. Also I have registered the models in admin.py.

Show all users related to tenant in one place

I'm working on an application in the saas model (django 2.1). I use the django tenants plugin (https://github.com/tomturner/django-tenants).
My problem is to display all tenants in the "public" schema. In such a way as to see how much each tenant has users, to be able to manage them, etc.
Is it a good architectural solution to put a foreign key into Tenant in the User model and save this column during the registration process?
Is there another way to do it?
Below is example, pseudo code:
class Tenant(TenantMixin):
name = models.CharField(_('Name of company'), max_length=50, unique=True)
on_trial = models.BooleanField(default=True)
paid_until = models.DateTimeField()
created = models.DateTimeField(auto_now_add=True)
updated = models.DateTimeField(auto_now=True)
auto_create_schema = True
def __str__(self):
return self.name
def get_name(self):
return self.name
class Domain(DomainMixin):
pass
class User(AbstractBaseUser, PermissionsMixin):
email = models.EmailField(_('Email address'), unique=True)
first_name = models.CharField(_('First name'), max_length=60, blank=True)
last_name = models.CharField(_('Last name'), max_length=60, blank=True)
member_of_company = models.ForeignKey(Tenant, on_delete=models.CASCADE, related_name='users', null=True, blank=True)
You can iterate over your tenants and get the Users for each tenant with the following code:
from django.db import connection
from django.contrib.auth import get_user_model
from tenant_schemas.utils import get_tenant_model
UserModel = get_user_model()
TenantModel = get_tenant_model()
for tenant in TenantModel.objects.all():
connection.set_tenant(tenant)
tenant_users = UserModel.objects.all()
# Do whatever you want with your users

Show InlineModelAdmin for nested Model

I have nested models with OneToOneFields and want to have an InlineModelAdmin form in one ModelAdmin to point to a nested model...
models.py:
class User(models.Model):
username = models.CharField(max_length=128)
password = models.charField(max_length=128)
class IdentityProof(models.Model):
user = models.OneToOneField(User, on_delete=models.PROTECT, related_name='proof')
proof_identity = models.FileField(upload_to='uploads/%Y/%m/%d/')
class Company(models.Model):
user = models.OneToOneField(User, on_delete=models.PROTECT, related_name='company')
name = models.CharField(max_length=128)
class Person(models.Model):
user = models.OneToOneField(User, on_delete=models.PROTECT, related_name='person')
name = models.CharField(max_length=128)
admin.py:
class IdentityProofInline(admin.TabularInline):
model = IdentityProof
#admin.register(Company)
class CompanyAdmin(admin.ModelAdmin):
inlines = [IdentityProofInline]
#admin.register(Person)
class PersonAdmin(admin.ModelAdmin):
inlines = [IdentityProofInline]
For CompanyAdmin or PersonAdmin, I want to show its User's IdentityProof. How can I do that ?
I tried to use fk_name = 'user_proof or other combinations but it doesn't work...
Thanks.
This is not possible without an external package.
See Django Nested Inline and Django Nested Admin

Use of foreignKey between app models in Django 1.10

I have a website built in Django 1.10. The site has 3 different apps: teams, members and news.
The first app, called teams has one model called Team.
This is the Team/models.py:
from django.db import models
from django.db.models.signals import pre_save
from django.utils.text import slugify
class Team(models.Model):
name = models.CharField(max_length=255)
description = models.TextField()
slug = models.CharField(max_length=255, default='team', editable=True)
class Meta:
ordering = ('name',)
def __unicode__(self):
return self.name
The second app, called members has one model called Member.
This is the Member/models.py:
from django.db import models
class Piloto(models.Model):
name = models.CharField(max_length=255)
biography = models.TextField()
slug = models.CharField(max_length=255, default='piloto', editable=True)
class Meta:
ordering = ('name',)
def __unicode__(self):
return self.name
What I want is include the name of the team inside the member profile, so I know it should be something like:
team_of_member = models.ForeignKey();
But I don't know what to put in the parenthesis or how to import the model of the team to the model of the member. I was following the documentation of Django 1.10 but it wasn't working, also I've tried this link but it doesn't work. Could you give a hand? Thanks
Edit:
I tried to do as #Bulva was suggesting, so my code is now like this:
from django.db import models
from equipos.models import Team
class Member(models.Model):
name = models.CharField(max_length=255)
team = models.ForeignKey('teams.Team', null=True)
biography = models.TextField()
slug = models.CharField(max_length=255, default='piloto', editable=True)
class Meta:
ordering = ('name',)
def __unicode__(self):
return self.name
What you want is to provide the teams the member is a member of. It all depends on your business logic. A team has many members by definition, but can a member be a part of many teams? If yes, you have a many-to-many relationship. If no, you have a one-to-many relationship.
Under one-to-many assumption, the foreignkey information has to be put in the referenced model. Then:
from django.db import models
from team.models import Team # Generally, apps are in all lower-case (assuming your app is called team)
class Member(models.Model):
name = models.CharField(max_length=255)
team = models.ForeignKey('team.Team', related_name = 'members', null=True) # Do not forget to put team.Team inside a pair of single-quotes.
biography = models.TextField()
slug = models.CharField(max_length=255, default='piloto', editable=True)
class Meta:
ordering = ('name',)
def __unicode__(self): # use def __str__(self): in Python 3+
return self.name
In your view, you can then say this:
albert_team = albert.team
albert_teammates = albert_team.members
Under Many-to-Many assumption, it is more natural to capture the relationship in the team model:
from django.db import models
from django.db.models.signals import pre_save
from django.utils.text import slugify
class Team(models.Model):
name = models.CharField(max_length=255)
description = models.TextField()
slug = models.CharField(max_length=255, default='team', editable=True)
members = models.ManyToManyField('team.Member', related_name = 'teams')
class Meta:
ordering = ('name',)
def __unicode__(self):
return self.name
In views.py:
albert_teams = albert.teams
all_albert_teammates = []
for team in albert_teams:
all_albert_teammates.append(team.members)
Try this:
from teams.models import Team
# in the member model field
team_of_member = models.ForeignKey(Team);
In order to do what you want (assuming the code provided is your full "problematic" code) you should:
In Member/models.py:
from django.db import models
from teams.models import Team # <--add this line
class Piloto(models.Model):
name = models.CharField(max_length=255)
team_of_member = models.ForeignKey(Team, on_delete=models.CASCADE); # <--add this line
biography = models.TextField()
slug = models.CharField(max_length=255, default='piloto', editable=True)
class Meta:
ordering = ('name',)
def __unicode__(self):
return self.name
After you do this, remember to:
python manage.py makemigrations
python manage.py migrate
to change your models state.
Good luck :)

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