There's a lot of questions worded similarly, but every single one I've seen is somebody trying to get some kind of data through a ManyToMany relationship before saving it. I'm not trying to use the relationship at all before saving, I just want to see if the user put anything there or not.
My model has a ForeignKey field pointing to a parent model, and two ManyToMany fields pointing to other models, but I only want users to be able to use one M2M field or the other, not both. This model is being edited through the admin as an inline on its parent.
models.py
class ProductSpecial(models.Model):
# name, slug, image, etc
class ProductSpecialAmount(models.Model):
special = models.ForeignKey(ProductSpecial, related_name='amounts', on_delete=models.CASCADE)
amount = models.IntegerField()
brands = models.ManyToManyField(ProductBrand, related_name='specials', blank=True)
lines = models.ManyToManyField(ProductLine, related_name='specials', blank=True)
admin.py
class ProductSpecialAmountInline(admin.StackedInline):
model = ProductSpecialAmount
# fieldsets, etc
#admin.register(ProductSpecial)
class ProductSpecialAdmin(admin.ModelAdmin):
inlines = [ProductSpecialAmountInline]
# fieldsets, etc
I only want users to be able to choose from brands or lines, but not both, and I would like to validate this before save and throw a validation error if necessary. My initial attempt was to just do...
class ProductSpecialAmount(models.Model):
# ...
def clean(self):
if self.brands and self.lines:
raise ValidationError('Please choose either brands or lines, not both', code='invalid')
...but that throws ValueError: "<ProductSpecialAmount: ProductSpecialAmount object (None)>" needs to have a value for field "id" before this many-to-many relationship can be used.
I get that I can't actually query the related ProductBrand or ProductModel objects before this object is saved, but I don't actually want any data from those objects right now, I just want to know if the user left either of the fields blank or not, and am wondering if that's possible to see at the model level.
Whether you actually want to use the data from a field or just see if it is blank, the problem is caused by referencing the m2m field in any way before saving the object. I had a similar problem which I fixed using a custom form as per: https://stackoverflow.com/a/7986937/19837155
This might be more difficult when you're using inlines, but it may be the easiest way to solve your problem.
Let us say I have a model which contains related (foreign key) fields. Likewise, those Foreign Key fields may refer to models which may or may not contain related fields. Note that relational fields in Django may be one-to-one, many-to-one, or many-to-many.
Now, given an instance of a model, I want to recursively and dynamically get all instances of the models related to it, either directly or indirectly down the line. Conceptually, i want to perform a traversal of the related objects and return them.
Example:
class Model1{
rfield1 = models.ForeignKey("Model2")
rfield2 = models.ManyToManyField("Model3")
normalfield1 = models.Charfield(max_length=50)
}
class Model2{
sfield = models.ForeignKey("Model3")
normalfield = models.CharField(max_length=50)
}
class Model3{
normalfield = models.CharField(max_length=50)
}
Let's say, I have an instance of model Model1 model1, and I want to get objects directly related to it i.e. all Model2 and Model3 objects, and also those which are indirectly related i.e. all Model3 objects related to the Model2 objects retrieved previously. I also want to consider the case of a One-to-One field where the related field is defined on the OTHER MODEL.
Also, note that it might not be the case that I know the model of an instance I'm currently working on. Let's say in the previous example, I may not now that model1 is an instance of Model1 model. So I want to perform all these dynamically.
In order to this, I think I need a way to get all related fields of an object.
How to get all the related fields of an object?
And how should I use them to get the actual related objects?
Or is there a way to better to do this? Thank you very much!
UPDATE:
I already know how to perform 1, and 2 basically follows directly from 1. :) Update later.
If you have model1 getting all it's many to many field names (etc) is easy since this is well know and these are all stored in the meta's 'local_many_to_many' list:
[field.name for field in model1._meta.local_many_to_many]
The foreign keys are a bit more tricky since they are stored with all other fields in the meta's 'local_fields' list. Hence we need to make sure that it has a relation of sorts. This can be done as follows:
[field.name for field in model1._meta.local_fields if field.rel]
This method has requires no knowledge of your models. Also further interrogation can be done on the field object if the name is not enough.
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.
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.
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