I want to create database model in django and define a foreign key. All the example I see in internet have following syntax.
attr = models.ForeignKey(Entity)
However I want a foreign key to be linked with single attribute from another entity and not the whole entity.
I have following models:
class User(models.Model):
user_id = models.PositiveIntegerField(primary_key=True)
password = models.CharField(max_length=20)
email = models.EmailField()
........
class ContentItem(models.Model):
content_id = models.PositiveIntegerField(primary_key=True)
title = models.CharField(max_length=100)
description = models.TextField()
.........
author_id = models.ForeignKey(User, on_delete= models.CASCADE)
Here, I want ContentItem.author_id to be foreign key for User.user_id and not the whole User model. While entering value for author_id, I simply want to enter plain id (1,2,3) and not all the instance of User
Sorry if the question is very general and thanks in advance
You should name your fk field like:
author = models.ForeignKey(User, on_delete= models.CASCADE)
because Django will create the attribute author_id automagically and
content_item.author_id
will give you access to the id of the user directly without an extra db hit. With your naming, you can access the id directly through author_id_id.
You have to override the str(self) method inside your models class.
Ex.
models.py
class User(models.Model):
.
.
.
def __str__(self):
return fâ{self.attribute-name}
Where attribute-name is name of attribute you want to reference.
Enjoy the coding đđ
Related
We have two Django models:
class Project(models.Model):
project_title = models.CharField(max_length=30)
owner = models.ForeignKey(User, null=True, on_delete=models.DO_NOTHING)
class User(models.Model):
usernmae = models.CharField(max_length=50)
active_project = models.ForeignKey(User, null=True, on_delete=models.DO_NOTHING, related_name='current_project')
I have a user with object (with id say 692). And this user created a project with id=12345, therefore these owner field will get have this particular referenced.
I want to delete that user. But it shows error that
delete on table "app_user" violates foreign key constraint
This is expected as on_delete=models.DO_NOTHING, was set. One way I found out was using on_delete=models.CASCADE.
Question: How should I go about deleting the user (692) without changing the model definition(having to re-run migration)?
Doing it manually by deleting the project first, leads to the same foreign-key error, as owner field is User object.
How to handle this mutual foreign key relationship while deleting, as deleting any one of those two throws the foreign-key exception?
Update
Some correction in the model definition username is the field name instead of usernmae (typo). And the foreignkey for project is Project not the User model.
class Project(models.Model):
project_title = models.CharField(max_length=30)
owner = models.ForeignKey(User, null=True, on_delete=models.DO_NOTHING)
class User(models.Model):
username = models.CharField(max_length=50)
active_project = models.ForeignKey(Project, null=True, on_delete=models.DO_NOTHING, related_name='current_project')
IF you really don't want to make a migration (any specific reason?) and if you are ok with doing this manually this time. Then you have two options:
Go into the admin panel and manually change the User field in the project instance to a different user or to NULL. Now you should be able to delete the User instance since it's not referred anymore into the project.
If that worked, you can then delete the project instane as well.
Curios if this will work, let me know!
I want to achieve a functionality, where I need to select a django model (e.g from a drop down list), and after selecting one, all the objects of that model shows up.
class Thread(models.Model):
sender = models.(???) # This need to be a field that can store a different model on a run time.
receiver = models.(???) # same here.
Is there any way that I can dynamically first select the model and then pick an object of that list. I have seen this functionality in odoo. But is there anything in Django?
Use Inheritance for in your Model and map your foreign key to User, and then pass either a teacher of student object.
You can use the many-to-many filed with multiple available choices of "Student" and "Teacher" from another Model.
class UserRole(models.Model):
STUDENT = 'STUDENT'
TEACHER = 'TEACHER'
ROLE_CHOICES = (
(STUDENT, 'student'),
(TEACHER, 'teacher'),
)
role_name = models.CharField(max_length=255, choices=ROLE_CHOICES)
def __str__(self):
return "{}".format(self.role_name)
class User(AbstractUser):
username = models.CharField(max_length=50, unique=True)
email = models.EmailField(_('email address'))
role = models.ManyToManyField(UserRole)
Class Thread(models.Model):
sender = models.OneToOneField(User, on_delete=models.CASCADE)
receiver = models.OneToOneField(User, on_delete=models.CASCADE)
This way you can only put available roles in sender and receiver fields of Thread.
The solution was possible with ajax too, but there also is another way in django which I was searching for.
class Test(models.Model):
content_type = models.ForeignKey(ContentType, on_delete=models.CASCADE)
object_id = models.PositiveIntegerField()
content_object = GenericForeignKey('content_type', 'object_id')
Have a good one.
Context: I'm forcing my self to learn django, I already wrote a small php based website, so I'm basically porting over the pages and functions to learn how django works.
I have 2 models
from django.db import models
class Site(models.Model):
name = models.CharField(max_length=50, unique=True)
def __str__(self):
return self.name
class Combo(models.Model):
username = models.CharField(max_length=50)
password = models.CharField(max_length=50)
dead = models.BooleanField(default=False)
timestamp = models.DateTimeField(auto_now_add=True)
siteID = models.ForeignKey(Site, on_delete=models.PROTECT)
class Meta:
unique_together = ('username','siteID')
def __str__(self):
return f"{self.username}:{self.password}#{self.siteID.name}"
When creating a view, I want to get the Combo objects, but I want to sort them first by site name, then username.
I tried to create the view, but get errors about what fields I can order by Cannot resolve keyword 'Site' into field. Choices are: dead, id, password, siteID, siteID_id, timestamp, username
def current(request):
current = Combo.objects.filter(dead=False).order_by('Site__name','username')
return render(request, 'passwords/current.html',{'current':current})
Since I'm not necissarily entering the sites into the database in alphabetical order, ordering by siteID wouldn't be useful. Looking for some help to figure out how to return back the list of Combo objects ordered by the Site name object then the username.
You can order this by siteID__name:
def current(request):
current = Combo.objects.filter(dead=False).order_by('siteID__name','username')
return render(request, 'passwords/current.html',{'current':current})
since that is the name of the ForeignKey. But that being said, normally ForeignKeys are not given names that end with an ID, since Django already adds an _id suffix at the end for the database field.
Normally one uses:
class Combo(models.Model):
# âŚ
site = models.ForeignKey(Site, on_delete=models.PROTECT)
if you want to give the database column a different name, you can specify that with the db_column=⌠parameter [Django-doc]:
class Combo(models.Model):
# âŚ
site = models.ForeignKey(
Site,
on_delete=models.PROTECT,
db_column='siteID'
)
How should I give ForeignKey to model?
Now models.py has User&Item table like
from django.db import models
# Create your models here.
class User(models.Model):
user_id = models.CharField(max_length=200)
name_id = models.CharField(max_length=200)
regist_date = models.DateTimeField(auto_now=True)
class Item(models.Model):
user_id = models.CharField(max_length=200)
name = models.CharField(max_length=200)
item_name = models.CharField(max_length=200)
price = models.CharField(max_length=200)
I wanna treat user_id&name_id as Foreing key.User table is parent table,and Item is child one.I think user_id&name_id in Item should have ForeignKey like
class Item(models.Model):
user_id = models.ForeignKey()
name = models.ForeignKey()
However,how should I connect these 2 model is User is parent&Item is child ?How should I write it?
For sure, you have to read that.
Looks like One User -> Many Items.
Its ForeignKey and we have to set it in User model.
class Item(models.Model):
...
class User(models.Model):
...
item = models.ForeignKey(Item)
You can use something like
class Item(models.Model):
user = models.ForeignKey(User, related_name='items')
Accessing user_id from item will be
item.user.user_id
Accessing all items from a user will be
user.items.all()
It's a good idea to read the documentation on the ForeignKey field.
As for your question, you can connect the Item model to the User model like this:
class Item(models.Model):
user = models.ForeignKey(User, on_delete=models.CASCADE)
Again, it's a good idea to read the documentation and figure out what exactly you need, such as what to do when a User object is deleted (the on_delete part in my code).
I have the following code:
class Item(models.Model):
name = models.CharField(max_length=100)
keywords = models.CharField(max_length=255)
type = models.ForeignKey(Type)
class Meta:
abstract = True
class Variant(models.Model):
test_field = models.CharField(max_length=255)
class Product(Item):
price = models.DecimalField(decimal_places=2, max_digits=8,null=True, blank=True)
brand = models.ForeignKey(Brand)
variant = models.ForeignKey(Variant)
def get_fields(self):
return [(field.name, field.value_to_string(self)) for field in Product._meta.fields]
def __unicode__(self):
return self.name
Im using Grappelli.
I want my Product to have multiple Variations. Should I use a manytomanyfield?
I want to be able to add Variants to my Product directly in the Admin. Now I get an empty dropwdown with no variants(because they doesnt exists).
I thought Django did this automatically when u specified a Foreign Key?
How can I get the Variant fields to display directly on my Product page in edit?
I've read someting about inline fields in Admin?
Well, it's the other way around :)
1/ Place the foreign key field in your Variant, not in your Product (what you describe is actually a OneToMany relationship).
2/ Link the Variant to your Product in the relative ProductAdmin in admin.py as an inline (i.e VariantInline).
See the docs for further informations : https://docs.djangoproject.com/en/1.6/ref/contrib/admin/#inlinemodeladmin-objects
Hope this helps !
Regards,