I suppose that question as old as it possibly could be, but I still have a hope that maybe someone has found some magical solution.
Everything is simple, I have a database (postgresql) with small table as an example, with already existing field with Primary key:
enter image description here
My goal is to add an ID field as a second primary key.
My Model looks like this:
class CdcConn(models.Model):
id = models.BigAutoField(primary_key=True)
conn_type = models.CharField(max_length=128, blank=True, null=True)
conn_name = models.CharField(max_length=128)
class Meta:
managed = True
db_table = 'cdc_conn'
After migration the result output is:
django.db.utils.ProgrammingError: multiple primary keys for table "cdc_conn" are not allowed
Tried many options, unique_together, constraints, some additional modules such as django-viewflow, and so on.
The issue is about 15 years old, documentation says that composite pk are not allowed, but still, I can't believe that such powerful tool as Django just can't handle that simple operation.
Any thoughts would be highly appreciated.
Related
I would like to add database constraints to my model, that require at least one of its fields to be not-null. When checking the m2m field, I get a FieldError: Cannot resolve keyword '' into field.
Is it possible to create such constraints?
Sample code:
class A(Model):
id = AutoField()
url = ManyToManyField(Url, blank=True)
description = TextField(null=True, blank=True)
class Meta:
constraints = [CheckConstraints(
check=(Q(description__isnull=False) | Q(url__isnull=False))),
name="someName"
)]
It is not possible to achieve this using the CheckConstraint functionality. Django translates all ORM commands to the low level DB specific commands, and such constraint creation isn't possible even on the DB level. In fact, we can apply CheckConstraint to a single row beeing added/updated only.
See full answer here:
https://stackoverflow.com/a/60799459/15090285
I want to have two versions of the same model while still benefiting from OneToOneField's reverse relation.
For example, let's say that I have the following models:
class Company(models.Model):
exists = models.BooleanField()
class ExtraInforation(models.Model):
company = models.OneToOneField(Company)
wealthy = models.BooleanField()
At this point my code uses the brilliance of the OneToOneField reverse relation, doing company.extrainformationcalls all over the place.
Then I get a new requirement: we can't trust the ExtraInformation without verifying it first! Pfft, any company could claim that it's wealty...
Any changes to ExtraInformation need to be confirmed before publishing. Let's say that the company isn't wealthy when it registers and that information gets confirmed. Later the company wants to mark itself wealthy. At that point there needs to be the confirmed/public version of ExtraInformation and the unconfirmed version that needs to be confirmed.
I want to be able to still keep those handy OneToOneField reverse relation calls but also have another version of the same data. The problem is, of course, that there can be only one row with reference to this company in the OneToOneField.
Currently my solution is to create a new table:
class ExtraInforationUnconfirmed(models.Model):
company = models.OneToOneField(Company)
wealthy = models.BooleanField()
Once the information is confirmed, the fields are copied from ExtraInforationUnconfirmed to ExtraInformation. This solution isn't very DRY or clean.
What would be the best way to solve this issue?
I studied proxy models and model inheritance. The best alternative way I could think of is to have a base model and inherit two models that have both have OneToOneField relation of their own to Company.
Add a boolean feild to the model and change it to true when confirmed:
class ExtraInforation(models.Model):
company = models.OneToOneField(Company)
wealthy = models.BooleanField()
confirmed = models.BooleanField(default=False)
UPDATE
Based on your comment I suggest a version filed which can be a simple integer or a datetime. I would avoid creating two models at any cost :)
class ExtraInforation(models.Model):
company = models.ForeignKey(Company, related_name='extrainformations')
wealthy = models.BooleanField()
# version = models.PositiveIntegerField()
version = models.DateTimeField(auto_now_add=True)
confirmed = models.BooleanField(default=False)
You can add a property to Company that returns the last extrainformation so that company.extrainformation will still work:
#property
def extrainformation(self):
return self.extrainformations.order_by("-version").first()
I've googled on and on, and I just don't seem to get it.
How do I recreate simple join queries in django?
in models.py (Fylker is county, Dagensrepresentanter is persons)
class Fylker(models.Model):
id = models.CharField(max_length=6, primary_key=True)
navn = models.CharField(max_length=300)
def __unicode__(self):
return self.navn
class Meta:
db_table = u'fylker'
class Dagensrepresentanter(models.Model):
id = models.CharField(max_length=33, primary_key=True)
etternavn = models.CharField(max_length=300, blank=True)
fornavn = models.CharField(max_length=300, blank=True)
fylke = models.ForeignKey(Fylker, db_column='id')
def __unicode__(self):
return u'%s %s' % (self.fornavn, self.etternavn)
class Meta:
ordering = ['etternavn'] # sette default ordering
db_table = u'dagensrepresentanter'
Since the models are auto-created by django, I have added the ForeignKey and tried to connect it to the county. The id fields are inherited from the db I'm trying to integrate into this django project.
By querying
Dagensrepresentanter.objects.all()
I get all the people, but without their county.
By querying
Dagensrepresentanter.objects.all().select_related()
I get a join on Dagensrepresentanter.id and Fylker.id, but I want thet join to be on fylke, aka
SELECT * FROM dagensrepresentanter d , fylker f WHERE d.fylke = f.id
This way I'd get the county name (Fylke navn) in the same resultset as all the persons.
Additional request:
I've read over the django docs and quite a few questions here at stackoverflow, but I can't seem to get my head around this ORM thing. It's the queries that hurt. Do you have any good resources (blogposts with experiences/explanations, etc.) for people accustomed to think of databases as an SQL-thing, that needs to start thinking in django ORM terms?
Your legacy database may not have foreign key constraints (for example, if it is using MyISAM then foreign keys aren't even supported).
You have two choices:
Add foreign key constraints to your tables (would involve upgrading to Innodb if you are on MyISAM). Then run ./manage inspectdb again and the relationships should appear.
Use the tables as is (i.e., with no explicit relationships between them) and compose queries manually (e.g., Mytable.objects.get(other_table_id=23)) either at the object level or through writing your own SQL queries. Either way, you lose much of the benefit of python's ORM query language.
I've searched around for a while, but can't seem to find an existing question for this (although it could be an issue of not knowing terminology).
I'm new to Django, and have been attempting to take a design which should be very expandable over time, and make it work with Django's ORM. Essentially, it's a series of many-to-many relationships using a shared junction table.
The design is a generic game crafting system, which says "if you meet [require], you can create [reward] using [cost] as materials." This allows items to be sold from any number of shops using the same system, and is generic enough to support a wide range of mechanics - I've seen it used successfully in the past.
Django doesn't support multiple M2M relationships sharing the same junction table (apparently since it has no way to work out the reverse relationship), so I seem to have these options:
Let it create its own junction tables, which ends up being six or more, or
Use foreign keys to the junction table in place of a built-in MTM relationship.
The first option is a bit of a mess, since I know I'll eventually have to add additional fields into the junction tables. The second option works pretty well. Unfortunately, because there is no foreign key from the junction table BACK to each of the other tables, I'm constantly fighting the admin system to get it to do what I want.
Here are the affected models:
class Craft(models.Model):
name = models.CharField(max_length=30)
description = models.CharField(max_length=300, blank=True)
cost = models.ForeignKey('Container', related_name="craft_cost")
reward = models.ForeignKey('Container', related_name="craft_reward")
require = models.ForeignKey('Container', related_name="craft_require")
class ShopContent(models.Model):
shopId = models.ForeignKey(Shop)
cost = models.ForeignKey('Container', related_name="shop_cost")
reward = models.ForeignKey('Container', related_name="shop_reward")
require = models.ForeignKey('Container', related_name="shop_require")
description = models.CharField(max_length=300)
class Container(models.Model):
name = models.CharField(max_length=30)
class ContainerContent(models.Model):
containerId = models.ForeignKey(Container, verbose_name="Container")
itemId = models.ForeignKey(Item, verbose_name="Item")
itemMin = models.PositiveSmallIntegerField(verbose_name=u"min amount")
itemMax = models.PositiveSmallIntegerField(verbose_name=u"max amount")
weight = models.PositiveSmallIntegerField(null=True, blank=True)
optionGroup = models.PositiveSmallIntegerField(null=True, blank=True,
verbose_name=u"option group")
Is there a simpler, likely obvious way to get this working? I'm attempting to allow inline editing of ContainerContent information from each related column on the Craft edit interface.
It sounds like you have a sort of "Transaction" that has a name, description, and type, and defines a cost, reward, and requirement. You should define that as a single model, not multiple ones (ShopContent, Craft, etc.).
class Transaction(models.Model):
TYPE_CHOICES = (('Craft', 0),
('Purchase', 1),
)
name = models.CharField(max_length=30)
description = models.CharField(max_length=300, blank=True)
cost = models.ForeignKey('Container')
reward = models.ForeignKey('Container')
require = models.ForeignKey('Container')
type = models.IntegerField(choices = TYPE_CHOICES)
Now Shop etc. can have a single ManyToManyField to Transaction.
Whether or not you use this particular model, the cost, reward and require relationships should all be in one place -- as above, or in OneToOne relationships with Craft, ShopContent etc. As you guessed, you shouldn't have a whole bunch of complex Many-To-Many through tables that are all really the same.
You mention at the bottom of your post that you're
attempting to allow inline editing of ContainerContent information from each related column on the Craft edit interface.
If you're modeling several levels of relationship, and using the admin app, you'll need to either apply some sort of nested inline patch, or use some sort of linking scheme like the one I use in my recent question, How do I add a link from the Django admin page of one object to the admin page of a related object?
I am smelling something is too complicated here, but I might be wrong. As a start,
is this any better? (ContainerContent will be figured out later)
class Cost(models.Model):
name = models.CharField(max_length=30)
class Reward(models.Model):
name = models.CharField(max_length=30)
class Require(models.Model):
name = models.CharField(max_length=30)
class Craft(models.Model):
name = models.CharField(max_length=30)
description = models.CharField(max_length=300, blank=True)
cost = models.ForeignKey(Cost)
reward = models.ForeignKey(Reward)
require = models.ForeignKey(Require)
class Shop(models.Model):
name = models.CharField(max_length=30)
crafts = models.ManyToMany(Craft, blank=True)
I'm not so great with databases so sorry if I don't describe this very well...
I have an existing Oracle database which describes an algorithim catalogue.
There are two tables algorithims and xref_alg.
Algorithims can have parents and children algorithms. Alg_Xref contains these relationships with two foreign keys - xref_alg and xref_parent.
These are the Django models I have so far from the inspectdb command
class Algorithms(models.Model):
alg_id = models.AutoField(primary_key=True)
alg_name = models.CharField(max_length=100, blank=True)
alg_description = models.CharField(max_length=1000, blank=True)
alg_tags = models.CharField(max_length=100, blank=True)
alg_status = models.CharField(max_length=1, blank=True)
...
class Meta:
db_table = u'algorithms'
class AlgXref(models.Model):
xref_alg = models.ForeignKey(Algorithms, related_name='algxref_alg' ,null=True, blank=True)
xref_parent = models.ForeignKey(Algorithms, related_name='algxref_parent', null=True, blank=True)
class Meta:
db_table = u'alg_xref'
On trying to query AlgXref I encounter this:
DatabaseError: ORA-00904: "ALG_XREF"."ID": invalid identifier
So the error seems to be that it looks for a primary key ID which isn't in the table.. I could create one but seems a bit pointless. Is there anyway to get around this? Or change my models?
EDIT: So after a bit of searching it seems that Django requires a model to have a primary key. Life is too short so have just added a primary key. Will this have any impact on performance?
This is currently a limitation of the ORM provided by Django. Each model has to have one field marked as primary_key=True, if there isn't one, the framework automatically creates an AutoField with name id.
However, this is being worked on as we speak as part of this year's Google Summer of Code and hopefully will be in Django by the end of this year. For now you can try to use the fork of Django available at https://github.com/koniiiik/django which contains an implementation (which is not yet complete but should be sufficient for your purposes).
As for whether there is any benefit or not, that depends. It certainly makes the database more reusable and causes less headaches if you just add an auto incrementing id column to each table. The performance impact shouldn't be too high, the only thing you might notice is that if you have a many-to-many table like this, containing only two ForeignKey columns, adding a third one will increase its size by one half. That should, however, be irrelevant as long as you don't store billions of rows in that table.