How to store a dictionary on a Django Model?

Question

I need to store some data in a Django model. These data are not equal to all instances of the model.

At first I thought about subclassing the model, but I’m trying t

Answer 1

This question is old, but I was having the same problem, ended here and the chosen answer couldn't solve my problem anymore.

If you want to store dictionaries in Django or REST Api, either to be used as objects in your front end, or because your data won't necessarily have the same structure, the solution I used can help you.

When saving the data in your API, use json.dump() method to be able to store it in a proper json format, as described in this question.

If you use this structure, your data will already be in the appropriate json format to be called in the front end with JSON.parse() in your ajax (or whatever) call.

Answer 2

As Ned answered, you won't be able to query "some data" if you use the dictionary approach.

If you still need to store dictionaries then the best approach, by far, is the PickleField class documented in Marty Alchin's new book Pro Django. This method uses Python class properties to pickle/unpickle a python object, only on demand, that is stored in a model field.

The basics of this approach is to use django's contibute_to_class method to dynamically add a new field to your model and uses getattr/setattr to do the serializing on demand.

One of the few online examples I could find that is similar is this definition of a JSONField.

Answer 3

If you are using Postgres, you can use an hstore field: https://docs.djangoproject.com/en/1.10/ref/contrib/postgres/fields/#hstorefield.

Answer 4

I came to this post by google's 4rth result to "django store object"

A little bit late, but django-picklefield looks like good solution to me.

Example from doc:

To use, just define a field in your model:

>>> from picklefield.fields import PickledObjectField
>>> class SomeObject(models.Model):
>>>     args = PickledObjectField()

and assign whatever you like (as long as it's picklable) to the field:

>>> obj = SomeObject()
>>> obj.args = ['fancy', {'objects': 'inside'}]
>>> obj.save()

Answer 5

Another clean and fast solution can be found here: https://github.com/bradjasper/django-jsonfield

For convenience I copied the simple instructions.

Install

pip install jsonfield

Usage

from django.db import models
from jsonfield import JSONField

class MyModel(models.Model):
    json = JSONField()

Answer 6

Think it over, and find the commonalities of each data set... then define your model. It may require the use of subclasses or not. Foreign keys representing commonalities aren't to be avoided, but encouraged when they make sense.

Stuffing random data into a SQL table is not smart, unless it's truly non-relational data. If that's the case, define your problem and we may be able to help.