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Is there a way to make the 'One' table optional in OneToManyRelationship in Django

I have two tables, one is Anonym the other is Userdatabase. I want my app to work without requiring any login info therefore it will work with Anonym only by using the deviceid of the user to process account information. If however, a user wants to access extra features they need to create a user with username/password. Then I will process the data using Userdatabase table. A user can have multiple devices so there is a OneToMany relationship in there, but a device doesn't have to have a User (they don't need to register) which breaks the relationship. Is there a way to make the Userdatabase table optional while keeping the OneToMany relationship? Perhaps by inserting a method or another class within UserDatabase? Please find the code below:
--Models--
class Anonym(models.Model):
deviceid=models.ForeignKey(Userdatabase,max_length=200,on_delete=models.SET_NULL,null=True)
accounttype=models.TextField(default='Free')
numberofattempts=models.IntegerField(default=0)
created=models.DateField(auto_now_add=True)
class Userdatabase(models.Model):
username=models.CharField(max_length=20,unique=True)
password=models.CharField(max_length=20)
deviceid=models.TextField(default='inputdeviceid')
accounttype=models.TextField(default='Free')
numberofattempts=models.IntegerField(default=0)
created=models.DateField(auto_now_add=True)
--urls--
urlpatterns=[path('deviceregister/<str:id>/',views.deviceregistration)]
--views--
def deviceregistration(request,id):
import time
deviceid=id
newdevice-models.Anonym(created=time.strftime("%Y-%m-%d"),deviceid=deviceid)
newdevice.save()
return HttpResponse('Succesful registration')
When I send a request as '/deviceregister/123456/' for example, django raises an ValueError saying Cannot assign "'123456'": "Anonym.deviceid" must be a "Userdatabase" instance.

you should search by fieldname, which contains id. in your case it is deviceid_id.
newdevice=models.Anonym(created=time.strftime("%Y-%m-%d"),deviceid_id=deviceid)
deviceid in your case should be Userdatabase.objects.get(pk=id)
deviceid=Userdatabase.objects.get(pk=id)
newdevice=models.Anonym(created=time.strftime("%Y-%m-%d"),deviceid=deviceid)
in my opinion - field names in your project really can confuse anyone

If you do not want to change your model, you can just link any newly-added device to a dummy user. When later a user want to link a device, replace dummy with the real user.
If you can change your model, you can remove the foreign key relationship, and add another table which links the id of both side: one field for deviceid and the other userid.
I know both options kind of smell, but at least they should work :)

Django multi-table inheritance - make sure only one child exists (CheckConstraint)

How can I make sure that a parent object has only one child/type?
class Property(...):
class Meta:
abstract = False
class Flat(Property):
pass
class House(Property):
pass
class Land(Property):
pass
I want every property object to have none or at most one child. It can be either flat, house or land (or null).
Is it possible to create a DB constraint for this?
My idea was to create a constraint that checks:
class Meta:
constraints = [
models.CheckConstraint(check=Q(Q(flat__isnull=True) & Q(house__isnull=True))
|
Q(Q(flat__isnull=True) & Q(land__isnull=True))
|
Q(Q(house__isnull=True) & Q(land__isnull=True)),
name="constraint")]
But apparently, there are no such fields on a DB level (you can get flat by property.flat getter in Django but not in DB)
Edit:
properties.Property: (models.E012) 'constraints' refers to the nonexistent field 'flat'.

But apparently, there are no such fields on a DB level (you can get flat by property.flat getter in Django but not in DB)
That is correct: Django adds a property to the Property model to lazily load the related Flat object and will make a query for that, but there is no database field named flat: this is just a query in reverse where Django basically queries with:
SELECT * FROM app_name_flat WHERE property_ptr=pk
with pk the primary key of the property object. It this makes a query.
A CHECK constraint [w3-schools] spans only over a row: it can not look on other rows nor can it look at other tables. It thus can not restrict other tables, and therefore is limited. It can for example prevent one column to have a certain value based on a value for another column in the same row (record), but that is how far a CHECK constraint normally looks.

Get the name of the collection which is currently using to store the documents

From the documentation of MongoEngine I have found that user can set preferred collection name.
class Users(Document):
user_id = SequenceField(primary_key=True)
meta = {'collection': "test_collection"}
Is there a way where I can get this collection name (here test_collection) later by sending query? My purpose is to do something after checking the existance of a collection.
The solution I have found that after getting the db name and list of collections checking the existance but here the limitation is I have to write by myself the required collection name.
db = get_db() # This one from connect parameter get the set database name
col = db.list_collection_names() # get all collections name of used database from the previous line
print("existance checking: ", "test_collection" in db.list_collection_names()) # here I don't want to hard coded the collection name "test_collection"
I have tried the following but get error
for x in Users.objects:
print("meta: ", x.meta)
Error_message:
AttributeError: 'Users' object has no attribute 'meta'
Reason of trying the former approach as I have got valid result from
for x in Users.objects:
print("user_id: ", x.user_id)
I am new in MongoDB so don't know how it works or whether it possible or not. So, if there is any query to acheive the current collection name specified in the Users class by PyMongo/ MongoEngine would be really helpful.

How to get child model related data in parent model query

I have two models:
class BusinessCard(models.Model):
name = models.CharField(_("name"),null=True,max_length=50)
class Contacts(models.Model):
businesscard_id = models.OneToOneField(BusinessCard,null=True,blank=True,related_name='contact_detail',db_column="businesscard_id")
bcard_json_data = JsonField(null=True)
I just want access contacts model data using business card model:
target_bacard=BusinessCard.objects.filter(id=target_bacard_id).select_related()
When we access the target_bacard.contact_detail it gives key errors.
How can I get the contacts data using target_bacard queryset.

use get() instead of filter() like:
target_bacard = BusinessCard.objects.get(id=target_bacard_id)
target_bacard.contact_detail

If you want to access the Contacts instance that is in the 1-to-1 relationship with a BusinessCard instance bacard, use the related name you specified in Contacts:
contact = bacard.contact_detail
Also, you have some misleading names: Contacts should rather be Contact since an instance of this model represents only one contact. And its field businesscard_id would better be named businesscard (note that the table column will be called businesscard_id at the database level automatically in that case and store the id of the related businesssscard) because in the ORM you get a BusinessCard model instance when you access it, and not just its id.

You have not passed related model (field) argument to select_related()
target_bacard=BusinessCard.objects.filter(id=target_bacard_id).select_related('contact_detail')
Assuming id of BusinessCard is unique, you may want to use ...objects.get(id=target_bacard_id) inplace of ...objects.filter(id=target_bacard_id). Anyway select_related() will work on both ways.
select_related() is used for saving database query.
here is the documentation

SQLAlchemy Self Referencing Relationship - NoReferencedTableError

I'm using SA 0.6.6, Declarative style, against Postgres 8.3, to map Python objects to a database. I have a table that is self referencing and I'm trying to make a relationship property for it's children. No matter what I try, I end up with a NoReferencedTableError.
My code looks exactly like the sample code from the SA website for how to do this very thing.
Here's the class.
class FilterFolder(Base):
__tablename__ = 'FilterFolder'
id = Column(Integer,primary_key=True)
name = Column(String)
isShared = Column(Boolean,default=False)
isGlobal = Column(Boolean,default=False)
parentFolderId = Column(Integer,ForeignKey('FilterFolder.id'))
childFolders = relationship("FilterFolder",
backref=backref('parentFolder', remote_side=id)
)
Here's the error I get:
NoReferencedTableError: Foreign key assocated with column 'FilterFolder.parentFolderId' could not find table 'FilterFolder' with which to generate a foreign key to target column 'id'
Any ideas what I'm doing wrong here?

This was a foolish mistake on my part. I typically specify my FK's by specifying the Entity type, not the string. I am using different schemas, so when defining the FK entity as a string I also need the schema.
Broken:
parentFolderId = Column(Integer,ForeignKey('FilterFolder.id'))
Fixed:
parentFolderId = Column(Integer,ForeignKey('SchemaName.FilterFolder.id'))

I checked your code with SQLAlchemy 0.6.6 and sqlite. I was able to create the tables, add a parent and child combination, and retrieve them again using a session.query.
As far as I can tell, the exception you mentioned (NoReferencedTableError) is thrown in schema.py (in the SQLAlchemy source) exclusively, and is not database specific.
Some questions: Do you see the same bug if you use an sqlite URL instead of the Postgres one? How are you creating your schema? Mine looks something like this:
engine = create_engine(db_url)
FilterFolder.metadata.create_all(self.dbengine)