Non-model field in Django model - django

I would like to have a model in Django that has multiple pictures associated with it. I'm evaluating possible options.
One picture for one model is easily done with the models.ImageField(...).
However, I would like a array (or set) of pictures. It can be just paths, not necessarily ImageField objects.
The problem is, how do I create that field in a Django model? I am assuming I will need to create a field that is not part of models.WhateverField. Is that possible? Can I define a non-model field, such as:
class MyModel:
name = models.CharField(max_length=10)
picture_list = []
and then do:
def sample_add_picture_view(request):
picture = "sample.jpg"
model = MyModel.objects.get(id=sample_id)
model.picture_list.append(picture)
model.save()
return HttpResponseRedirect('index.html')
Could this be done? If not, what could be a better solution? Thank you !

You need to create two separate models and link them with a ForeignKey field, like so:
class Item(models.Model):
name = models.CharField(max_length=255)
class ItemImage(models.Model):
image = models.ImageField(upload_to="item_images")
item = models.ForeignKey('Item', related_name="images")
It is possible to make a custom field to store multiple items, but it's a really bad idea. You would have to serialise an array into the database, making maintenance very difficult. Using a separate model means you can store extra information such as upload times, image captions etc with little extra effort.

Related

How best to validate number of total reverse relationships before saving

I have an Item model that is reverse-related to two other models (ItemComponent and or ItemComponentCategory). The idea is I'd like to be able to validate that Items have no more than 4 relations to the two other models, combined, before being saved.
class Item(models.Model):
name = models.CharField(max_length=40, unique=True)
class ItemComponent(models.Model):
parent_item = models.ForeignKey(Item, related_name='components')
class ItemComponentCategory(models.Model):
parent_item = models.ForeignKey(Item, related_name='categories')
I'd like to create a validation that raises an error before saving either the Item, ItemComponent, or ItemComponentCategory objects if the saved objects will result in > 4 object relations between them.
I have tried adding something like this to the clean methods for all three:
def clean(self):
if (self.parent_item.components.count() + self.parent_item.categories.count()) > 4:
raise ValidationError(_(f'Items can have no more than 4 components and/or component categories'))
This seems to work as long as the Items and their relations are already saved with 4 and you're trying to add more relations.
However, if I create a TabularInline in the ItemAdmin to add these 'sub types,' if you will.. I can create a new Item and add as many of these sub types and save it no problem.
What am I missing here?
It seems my issue is related to timing and order of processes.
When saving an admin page with inline references, the model of the main page is saved first then the inline objects. This poses an issue when trying to use a model's clean() method to validate against the relation, because it simply does not exist yet.
As mentioned in THIS article it seems that my solution is to modify the modelAdmin functions to ensure that the inline objects are saved first, then validate the main objects after they're saved.
Also, I think this save_related method for modelAdmin will come in handy.

Django One(Many)-To-Many Relation in reusable app

I have a model named Exam. each Exam has a set of users called participants. The only way I found to keep such set in Django is to add a field in User model. But I'd prefer to write this model to be as independent as possible so if later I want to use it again I can do it without changing my User model. So How can I handle having such set without manually modifying the User model fields?
Regarding your comment here is what you could do something like this:
class Exam(models.Model):
participants = models.ManyToMany(settings.AUTH_USER_MODEL, through='Participation')
class Participation(models.Model)
user = models.OneToOneField(settings.AUTH_USER_MODEL)
exam = models.ForeignKey('Exam')
active = models.BooleanField(default=False)
Another option would be to use Django's limit_coices_to. It's not transaction-save, but might do the job. You would just limit to choices to all non-related objects.

Building a list of ModelForms based on relation

I think this is best explained with some simple model code (I'm writing this from scratch so possible syntax issues - unimportant here):
class Car(models.Model)
make = models.CharField(...)
model = models.CharField(...)
class StatisticType(models.Model):
name = models.CharField(...)
class Statistic(models.Model)
car = models.ForeignKey('Car')
stype = models.ForeignKey('StatisticType')
data = models.CharField(...)
class Meta:
unique_together = (('car', 'stype'),)
We have a car with some hard-coded stats and we have some database controlled statistics. I might add Colours, Wheel Size, etc. The point is it's editable from the admin so neither I or the client need to climb through the data, but it's limited so users can only pick one of each stat (you can't define "Colours" twice).
So I'm trying to write the data input form for this now and I want a list of optional ModelForms that I can chuck on the page. I've got the simplest ModelForm possible:
class StatisticForm(forms.ModelForm):
class Meta:
model = Statistic
The tricky part (in my head) is generating an instance of this ModelForm for each StatisticType, regardless of it existing yet. That is to say if a Car object doesn't have a Colour assigned to it, the form still shows. Similarly, if it does, that instance of a Statistic is loaded in the ModelForm.
In my view, how do I generate a list of these things, regardless of there being a pre-existing instance of any given Statistic?
This seems like it should be a stupidly simple thing to do but it's late on Friday and everything looks skwonky.
Sounds like you might want to leverage an inline model formset factory.
That would allow you to create as many instances of your Statistic object as you need. If you're needing to create instances of your StatisticType on the fly, that's a bit different.
When Django instantiates forms, for a foreign key, m2m or choice field, it will only accept choices that it deems "valid", and will complain if you add a choice using JavaScript that doesn't exist in a related model or set of choices server-side.
So, if you need to make StatisticTypes on the fly, and then populate formset instances with this new value, I would suggest using Knockout.js. It's very good at keeping lots of DOM elements in sync when data changes.

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