Django making field related to other field (object) value - django

few years ego I worked with Odoo framework. and Odoo has very nice feature like this:
partner_id = field.Many2one(Partner)
partner_name = fields.Char(string='Partner name', related='partner_id.name')
basically whenever you would assign different partner_id from Partner table, partner_name would be assigned automatically. Now I started to work with django (absolute newbie), and I can't seem to find a similar functionality.
My question is what could be possible solution for this problem. Maybe there are already established external libraries that has this sort of functionality?
Expected result:
product = models.ForeignKey(Product)
product_color = models.CharField(string='Partner name', related='product.color')
having in mind that product object would have color field and it would be assigned to product_color whenever product field value Product object color value changes. Also what about storing it to database? Would be nice if there was an option to chose between storing it in database or getting it on the fly.
Cheers!

Creating a getter is pretty easy, because you can simply have functions in a Python object behave as a property:
class SampleModel(models.Model):
product = models.ForeignKey(Product)
#property
def product_color(self):
return self.product.color
This does retrieve the property on the fly, which will cause a call to the database.

Duplicating data, is usually a (more severe) antipattern. Synchronizing data, even in two tables in the same database, often turns out harder than one might expect. Even if you would use Django's signal framework for example, then some Django ORM calls can circumvent that (for example .update(..) [Django-doc]). But even if you somehow would cover those cases, then another program that talks to the database could update one of the two fields.
Most databases have triggers that can help. But again, the number of cases to cover are often larger than expected. For example, if the Product that we refer to is removed, then or the foreign key now points to a different Product, then we will need to update that field.
Therefore it is often better, to fetch the name of the related product when we need it. We can do so by (a) defining a property; or (b) make an annotation, for example in the manager.
Defining a property
We can define a property that will load the related product, and fetch the related name, like:
class Order(models.Model):
product = models.ForeignKey(Product, on_delete=models.PROTECT)
#property
def product_name(self):
return self.product.name
Then we can fetch the product name with some_order.product_name. This might not be very efficient if we need to fetch it often, since the relations are, by default, loaded lazily in Django, and thus can result in an N+1 problem.
Annotate the queryset
We can make an annotation that will fetch the name of the product in the same query when we fetch the Order, for example:
from django.db.models import F
class OrderManager(models.Manager):
def get_queryset(self):
return super().get_queryset().annotate(
product_name=F('product__name')
)
class Order(models.Model):
product = models.ForeignKey(Product, on_delete=models.PROTECT)
objects = OrderManager()
Then if we fetch an order. For example with Order.objects.get(pk=1), then that Order object will have an attribute product_name with the name of the product.

Related

Displaying all fields of Foreign Key Model

What I want is to retrieve all the fields belonging to a Model of a foreign key.
My models for example:
class BaseProduct(models.Model):
name = models.CharField(max_length=256)
variant = models.CharField(max_length=256, default='N/A')
type = models.ForeignKey(ProductType)
class ProductType(models.Model):
name = models.CharField(max_length=256,blank=False,null=False)
sofa = models.ForeignKey(SofaProduct, blank=True, null=True)
toaster = models.ForeignKey(ToasterProduct, blank=True, null=True)
These are just examples, there can be any number of ProductType models each with any number of fields.
In my template I can display all the fields of the BaseProduct by using the BaseProduct ID. What I want is to display all the fields of the FK.
For example if type = sofa in BaseProduct, I need to retrieve and display all sofa fields as well as BaseProduct fields.
(disclaimer: I have a tendency to give really long answers. You'll have to forgive me for that)
First rule of schema design - It should reflect your real world business logic (not the actual business action mind you, just the implications of the relationships). For example, if I have a class Person I can create a class Pet with a foreginKey to Person which translates to - every person can have multiple pets.
If we apply that logic to your schema we see that ProductType is a class that has a foreignKey to both Sofas and Toasters, which means each Toaster can have multiple Sofas and vice versa. Last time I checked, I never heard of a Sofa that had a Toaster.
In other words - you need to think what you're actually trying to achieve here. I'm guessing BaseProduct is a basic class that has common fields, and Sofa and Toaster are different types of products. Since they are different, they have their own special fields, and shouldn't be related, so it makes sense to have them as separate models. So why do you even need ProductType? To define the name Toaster? You're already defining an entire model! Why do you need to keep its name on a different table (and not, say, some custom method that always returns "I am a toaster, hear me roar")?
My best guess is that you want to be able to define new types of products on the go. However, if you intend to keep them separated on the model level, then you'll have to create a model for each new product. And if you want to be able to simple define a new model with ProductType, then you either need to have one Product class to manage them all, or you want a complicated dynamic system that can create new models on the fly.
Let's break those options down:
Create a generic product and a type class, like you did there:
class ProductType(models.Model):
name = models.CharField(max_length=256,blank=False,null=False)
class Product(models.Model):
name = models.CharField(max_length=256)
variant = models.CharField(max_length=256, default='N/A')
type = models.ForeignKey(ProductType)
Now each product can only be of one type, and you can always create new types on the go. This of course means all Product objects will share the same fields, and is very limiting. You won't have the same flexibility for each type like you would before (no sofa-only fields), but on the other hand it will be easier to create dynamic types of objects - you just define a new ProductType and bam you have a whole new group of products.
Create a basic abstract Product model, and define a new sub-model for each new type of product. You'll have a lot more flexibility for each one, but defining new types will always require defining a new model and setting up a table for it. With this scheme you don't need the ProductType object at all because the different models define the different types (there's no need for duplicity).
You can create some kind of admin page for the process, but it's not going
to be very easy to setup, and you might find yourself eventually with too many tables
(which can be especially problematic if you need to sometimes query
on all products - you'll have to join a lot of different tables,
which is not very efficient).
Use a non-relational database with some dynamic-models know how and disco*
*ok, it's actually more complicated than that, but the explanation on how to combine them is way too long, even for my answer. If it seems over your head, forget about it. If you have some idea about how non-relation databases work, you can probably figure it out yourself
Your question is somewhat unclear.
I think you want Django modal forms to display all fields of an modal.
def ListForm(Forms.form):
model = MyModel
fields='__all__' #Sets display all
fk_name ="Model_to_use" #Is needed when your model has more then one fk
Django model form
You can use _set for accessing related objects. For example, if you have two models like these:
class MyModel(models.Model):
name = models.CharField(max_length=200)
somedata = models.CharField(max_length=200)
class AnotherModel(models.Model):
name = models.CharField(max_length=256,blank=False,null=False)
referral = models.ForeignKey(MyModel)
type = models.CharField(max_length=256,blank=False,null=False)
you can access the name field of AnotherModel with
>>> m = MyModel.objects.get(id=1)
>>> m.AnotherModel_set.all()[0].name
See: https://docs.djangoproject.com/en/dev/topics/db/queries/#related-objects
On a side note, you should probably rethink your models structure, as yuvi pointed out.

How to prefetch aggregated #property in Django?

We have two models (simplified versions):
class Contestant(models.Model):
email = models.EmailField(max_length=255, unique=True)
# plus some other fields
#property
def total_points(self):
return self.points.aggregate(total=Sum('value'))['total'] or 0
class Points(models.Model):
contestant = models.ForeignKey(Contestant, related_name='points')
value = models.PositiveIntegerField()
# plus some other fields which determine based on what we
# awarded ``Points.value``
When we display a list of contestants along with their total_points value, it
results in an extra query for each result - i.e. the following queries are performed:
fetch list of contestants
fetch total_points value of 1st contestant
fetch total_points value of 2nd contestant
etc
I tried altering the queryset to prefetch the data as follows:
Contestant.objects.filter(...).prefetch_related('points')
..., however even though it works, the prefetched data is not utilized when
listing contestants (so each result still tries to fetch total_points
in a separate query).
Is it possible to:
somehow tell the ORM to use prefetched values for the #property field when
populating data for individual model objects (e.g. access the prefetched value inside the Contestant.total_points #property method)?
or to prefetch them in a different way (as opposed to the example above)?
or to use a completely different approach achieving the same result?
(I'm listing results in tastypie, if it matters.)
Thank you.
When your aim is to add aggregated values to each item, you should use annotate, instead of aggregate.
For example (a simple query, no additional methods required):
Contestant.objects.filter(...).annotate(total_points=Sum('points__value'))
If you really want to put this code out of your query: you can, but a model method is not a right way to do this. Methods on models are for operations on single instances. If you want to do something on a whole QuerySet use an ORM Manager instead.
With a Manager this would look like this:
class TotalPointsManager(models.Manager):
def get_queryset(self):
return super(TotalPointsManager, self).get_queryset().annotate(total_points=Sum('points__value'))
class Contestant(models.Model):
email = models.EmailField(max_length=255, unique=True)
objects = TotalPointsManager() # You are overriding the default manager!
and then you would construct your query as usual (you can drop prefetch_related):
Contestant.objects.filter(...)
...and total_points field would become "magically" available for every object.

Django and computed fields in a legacy database

I am gradually replacing a legacy database front end with a django based system. All models are Managed = False, to preserve the original db structure.
However I have struck a problem where a field in a table is computed. The field is defined in (pseudo) sql as full_name = fname|| ||lname.
I can define the full_name field as a charfield; and I can read it no problems, however any attempt to update the record results in an update error on that field.
I can use a #property; but that replicates the functionality in django, rather than displaying the results from the db itself. Using this approach will cause problems with more complex fields (in tables I am yet to get to) that are defined with UDFs.
What is really needed is a 'read_only' or 'computed' attribute on the model itself; what is the best way to achieve this?
Do you just want to define a method in your class? Something like:
def Person(models.Model):
fname=models.CharField(...)
lname=models.CharField(...)
def fullname(self):
return self.fname+" "+self.lname
(not exactly sure what Managed=False means...)
if you are trying to make calculation on a database models and pass the value of a model field to another model field of the same class model, using a defined function then this solution might help you. for example lets assume you have an investment company and you give 20% per month for the capital each user invested, you would want want to pass value from capital model to a function that calculates the percentage interest, and then you will pass that function into another field monthly_payment and get saved in the database.
1) pip install django-computed-property
2) add 'computed_property' to your installed apps in project settings.
3) in your models.py, import computed_property then
class Investment(models.Model):
name = models.CharField(max_length=200)
capital = models.FloatField(null=False)
percentage = models.CharField(max_length=5)
duration = models.CharField(max_length=10, default="1 months")
monthly_payment = computed_property.ComputedFloatField( compute_from='monthly_percentage', null=False, blank=False)
then your function to perform the calculation will go thus
#property
def monthly_percentage(self):
return (20 / 100) * self.capital
Note: what i discovered was if you use the inbuilt django fields be it FloatFiled or IntegerField, this function won't read the amount you are passing in to get your 20% calculations.i hope this works for you as i stated all as they worked for me,cheers.

Making a fairly complex Django model method sortable in admin?

I have a reasonably complex custom Django model method. It's visible in the admin interface, and I would now like to make it sortable in the admin interface too.
I've added admin_order_field as recommended in this previous question, but I don't fully understand what else I need to do.
class Book(models.Model):
id = models.IntegerField(primary_key=True)
title = models.CharField(max_length=200)
library_id = models.CharField(max_length=200, unique=True)
def current_owner(self):
latest_transaction = Transaction.objects.filter(book=self)[:1]
if latest_transaction:
if latest_transaction[0].transaction_type==0:
return latest_transaction[0].user.windows_id
return None
current_owner.admin_order_field = 'current_owner'
Currently, when I click on the current_owner field in the admin interface, Django gives me
FieldError at /admin/books/book/
Cannot resolve keyword 'current_owner' into field
Do I need to make a BookManager too? If so, what code should I use? This isn't a simple Count like the example in the previous question, so help would be appreciated :)
Thanks!
The Django admin won't order models by the result of a method or any other property that isn't a model field (i.e. a database column). The ordering must be done in the database query, to keep things simple and efficient.
The purpose of admin_order_field is to equate the ordering of a non-field property to the ordering of something that is a field.
For example, a valid values current_owner.admin_order_field could be id, title or library_id. Obviously none of these makes sense for your purpose.
One solution would be to denormalise and always store current_owner as a model field on Book; this could be done automatically using a signal.
You can't do this. admin_order_field has to be a field, not a method - it's meant for when you have a method that returns a custom representation of an underlying field, not when you do dynamic calculations to provide the value. Django's admin uses the ORM for sorting, and that can't sort on custom methods.

Creation of dynamic model fields in django

This is a problem concerning django.
I have a model say "Automobiles". This will have some basic fields like "Color","Vehicle Owner Name", "Vehicle Cost".
I want to provide a form where the user can add extra fields depending on the automobile that he is adding. For example, if the user is adding a "Car", he will extra fields in the form, dynamically at run time, like "Car Milage", "Cal Manufacturer".
Suppose if the user wants to add a "Truck", he will add "Load that can be carried", "Permit" etc.
How do I achieve this in django?
There are two questions here:
How to provide a form where the user can add new fields at run time?
How to add the fields to the database so that it can be retrieved/queried later?
There are a few approaches:
key/value model (easy, well supported)
JSON data in a TextField (easy, flexible, can't search/index easily)
Dynamic model definition (not so easy, many hidden problems)
It sounds like you want the last one, but I'm not sure it's the best for you. Django is very easy to change/update, if system admins want extra fields, just add them for them and use south to migrate. I don't like generic key/value database schemas, the whole point of a powerful framework like Django is that you can easily write and rewrite custom schemas without resorting to generic approaches.
If you must allow site users/administrators to directly define their data, I'm sure others will show you how to do the first two approaches above. The third approach is what you were asking for, and a bit more crazy, I'll show you how to do. I don't recommend using it in almost all cases, but sometimes it's appropriate.
Dynamic models
Once you know what to do, this is relatively straightforward. You'll need:
1 or 2 models to store the names and types of the fields
(optional) An abstract model to define common functionality for your (subclassed) dynamic models
A function to build (or rebuild) the dynamic model when needed
Code to build or update the database tables when fields are added/removed/renamed
1. Storing the model definition
This is up to you. I imagine you'll have a model CustomCarModel and CustomField to let the user/admin define and store the names and types of the fields you want. You don't have to mirror Django fields directly, you can make your own types that the user may understand better.
Use a forms.ModelForm with inline formsets to let the user build their custom class.
2. Abstract model
Again, this is straightforward, just create a base model with the common fields/methods for all your dynamic models. Make this model abstract.
3. Build a dynamic model
Define a function that takes the required information (maybe an instance of your class from #1) and produces a model class. This is a basic example:
from django.db.models.loading import cache
from django.db import models
def get_custom_car_model(car_model_definition):
""" Create a custom (dynamic) model class based on the given definition.
"""
# What's the name of your app?
_app_label = 'myapp'
# you need to come up with a unique table name
_db_table = 'dynamic_car_%d' % car_model_definition.pk
# you need to come up with a unique model name (used in model caching)
_model_name = "DynamicCar%d" % car_model_definition.pk
# Remove any exist model definition from Django's cache
try:
del cache.app_models[_app_label][_model_name.lower()]
except KeyError:
pass
# We'll build the class attributes here
attrs = {}
# Store a link to the definition for convenience
attrs['car_model_definition'] = car_model_definition
# Create the relevant meta information
class Meta:
app_label = _app_label
db_table = _db_table
managed = False
verbose_name = 'Dynamic Car %s' % car_model_definition
verbose_name_plural = 'Dynamic Cars for %s' % car_model_definition
ordering = ('my_field',)
attrs['__module__'] = 'path.to.your.apps.module'
attrs['Meta'] = Meta
# All of that was just getting the class ready, here is the magic
# Build your model by adding django database Field subclasses to the attrs dict
# What this looks like depends on how you store the users's definitions
# For now, I'll just make them all CharFields
for field in car_model_definition.fields.all():
attrs[field.name] = models.CharField(max_length=50, db_index=True)
# Create the new model class
model_class = type(_model_name, (CustomCarModelBase,), attrs)
return model_class
4. Code to update the database tables
The code above will generate a dynamic model for you, but won't create the database tables. I recommend using South for table manipulation. Here are a couple of functions, which you can connect to pre/post-save signals:
import logging
from south.db import db
from django.db import connection
def create_db_table(model_class):
""" Takes a Django model class and create a database table, if necessary.
"""
table_name = model_class._meta.db_table
if (connection.introspection.table_name_converter(table_name)
not in connection.introspection.table_names()):
fields = [(f.name, f) for f in model_class._meta.fields]
db.create_table(table_name, fields)
logging.debug("Creating table '%s'" % table_name)
def add_necessary_db_columns(model_class):
""" Creates new table or relevant columns as necessary based on the model_class.
No columns or data are renamed or removed.
XXX: May need tweaking if db_column != field.name
"""
# Create table if missing
create_db_table(model_class)
# Add field columns if missing
table_name = model_class._meta.db_table
fields = [(f.column, f) for f in model_class._meta.fields]
db_column_names = [row[0] for row in connection.introspection.get_table_description(connection.cursor(), table_name)]
for column_name, field in fields:
if column_name not in db_column_names:
logging.debug("Adding field '%s' to table '%s'" % (column_name, table_name))
db.add_column(table_name, column_name, field)
And there you have it! You can call get_custom_car_model() to deliver a django model, which you can use to do normal django queries:
CarModel = get_custom_car_model(my_definition)
CarModel.objects.all()
Problems
Your models are hidden from Django until the code creating them is run. You can however run get_custom_car_model for every instance of your definitions in the class_prepared signal for your definition model.
ForeignKeys/ManyToManyFields may not work (I haven't tried)
You will want to use Django's model cache so you don't have to run queries and create the model every time you want to use this. I've left this out above for simplicity
You can get your dynamic models into the admin, but you'll need to dynamically create the admin class as well, and register/reregister/unregister appropriately using signals.
Overview
If you're fine with the added complication and problems, enjoy! One it's running, it works exactly as expected thanks to Django and Python's flexibility. You can feed your model into Django's ModelForm to let the user edit their instances, and perform queries using the database's fields directly. If there is anything you don't understand in the above, you're probably best off not taking this approach (I've intentionally not explained what some of the concepts are for beginners). Keep it Simple!
I really don't think many people need this, but I have used it myself, where we had lots of data in the tables and really, really needed to let the users customise the columns, which changed rarely.
Database
Consider your database design once more.
You should think in terms of how those objects that you want to represent relate to each other in the real world and then try to generalize those relations as much as you can, (so instead of saying each truck has a permit, you say each vehicle has an attribute which can be either a permit, load amount or whatever).
So lets try it:
If you say you have a vehicle and each vehicle can have many user specified attributes consider the following models:
class Attribute(models.Model):
type = models.CharField()
value = models.CharField()
class Vehicle(models.Model):
attribute = models.ManyToMany(Attribute)
As noted before, this is a general idea which enables you to add as much attributes to each vehicle as you want.
If you want specific set of attributes to be available to the user you can use choices in the Attribute.type field.
ATTRIBUTE_CHOICES = (
(1, 'Permit'),
(2, 'Manufacturer'),
)
class Attribute(models.Model):
type = models.CharField(max_length=1, choices=ATTRIBUTE_CHOICES)
value = models.CharField()
Now, perhaps you would want each vehicle sort to have it's own set of available attributes. This can be done by adding yet another model and set foreign key relations from both Vehicle and Attribute models to it.
class VehicleType(models.Model):
name = models.CharField()
class Attribute(models.Model):
vehicle_type = models.ForeigngKey(VehicleType)
type = models.CharField()
value = models.CharField()
class Vehicle(models.Model):
vehicle_type = models.ForeigngKey(VehicleType)
attribute = models.ManyToMany(Attribute)
This way you have a clear picture of how each attribute relates to some vehicle.
Forms
Basically, with this database design, you would require two forms for adding objects into the database. Specifically a model form for a vehicle and a model formset for attributes. You could use jQuery to dynamically add more items on the Attribute formset.
Note
You could also separate Attribute class to AttributeType and AttributeValue so you don't have redundant attribute types stored in your database or if you want to limit the attribute choices for the user but keep the ability to add more types with Django admin site.
To be totally cool, you could use autocomplete on your form to suggest existing attribute types to the user.
Hint: learn more about database normalization.
Other solutions
As suggested in the previous answer by Stuart Marsh
On the other hand you could hard code your models for each vehicle type so that each vehicle type is represented by the subclass of the base vehicle and each subclass can have its own specific attributes but that solutions is not very flexible (if you require flexibility).
You could also keep JSON representation of additional object attributes in one database field but I am not sure this would be helpfull when querying attributes.
Here is my simple test in django shell- I just typed in and it seems work fine-
In [25]: attributes = {
"__module__": "lekhoni.models",
"name": models.CharField(max_length=100),
"address": models.CharField(max_length=100),
}
In [26]: Person = type('Person', (models.Model,), attributes)
In [27]: Person
Out[27]: class 'lekhoni.models.Person'
In [28]: p1= Person()
In [29]: p1.name= 'manir'
In [30]: p1.save()
In [31]: Person.objects.a
Person.objects.aggregate Person.objects.all Person.objects.annotate
In [32]: Person.objects.all()
Out[33]: [Person: Person object]
It seems very simple- not sure why it should not be a considered an option- Reflection is very common is other languages like C# or Java- Anyway I am very new to django things-
Are you talking about in a front end interface, or in the Django admin?
You can't create real fields on the fly like that without a lot of work under the hood. Each model and field in Django has an associated table and column in the database. To add new fields usually requires either raw sql, or migrations using South.
From a front end interface, you could create pseudo fields, and store them in a json format in a single model field.
For example, create an other_data text field in the model. Then allow users to create fields, and store them like {'userfield':'userdata','mileage':54}
But I think if you're using a finite class like vehicles, you would create a base model with the basic vehicle characteristics, and then create models that inherits from the base model for each of the vehicle types.
class base_vehicle(models.Model):
color = models.CharField()
owner_name = models.CharField()
cost = models.DecimalField()
class car(base_vehicle):
mileage = models.IntegerField(default=0)
etc