Convert JSON data into an instance of a Django model

Question

I need to convert JSON data to django model.

This is my JSON data

{
  "data": [
    {
      "id": "20ad5d9c-b32e-4599-8866-a3aaa5ac77de",
      "name": "name_1"
    },
    {
      "id": "7b6d76cc-86cd-40f8-be90-af6ced7fec44",
      "name": "name_2"
    },
    {
      "id": "b8843b1a-9eb0-499f-ba64-25e436f04c4b",
      "name": "name_3"
    }
  ]
}

This is my django method

def get_titles():
    url = 'http://localhost:8080/titles/' 
    r = requests.get(url)
    titles = r.json()
    print(titles['data'])

What I need is convert the JSON into an instance of the model and pass it to the template. Please let me know how to convert JSON to Model.

Laur Ivan · Accepted Answer · 2017-09-22 07:29:05Z

15

Using JSON in Django templates

You don't have to convert the JSON structure into a Django model just to use it in a Django template: JSON structures (Python dicts) work just fine in a Django template

e.g. if you pass in {'titles': titles['data']} as the context to your template, you can use it as:

{% for title in titles %}
    ID is {{title.id}}, and name is {{title.name}}
{% endfor %}

As long as you don't need to store the data with Django, the above solution works just fine. If you want to store, read below.

Make a model

You can create a model to store that JSON data in. Once stored you can pass the queryset to your template

class Title(models.Model)
    id = models.CharField(max_length=36)
    name = models.CharField(max_length=255)

or use an UUIDField

class Title(models.Model)
    id = models.UUIDField(primary_key=True)
    name = models.CharField(max_length=255)

Store the data in a Django model

# Read the JSON
titles = r.json()
# Create a Django model object for each object in the JSON 
for title in titles['data']:
    Title.objects.create(id=title['id'], name=title['name'])

Use stored data to pass as template context

# Then pass this dict below as the template context
context = {'titles': Title.objects.all()}

edited Sep 22, 2017 at 7:29

Laur Ivan

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answered Jan 16, 2016 at 10:50

bakkal

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6 Comments

Damith Ganegoda Over a year ago

Thanks you'r comment. I've an question. I didn't config the database. It just consume REST service. How do I create object and get all objects that context?

bakkal Over a year ago

Put the Title model I described in my answer, in a file inside your Django app called models.py, add your app to INSTALLED_APPS in your settings.py, run database migration to create the SQL tables etc, more on that here: docs.djangoproject.com/en/1.9/topics/db/models

bakkal Over a year ago

If want a model without a DB table, then you don't really need a model in the Django sense :) You are fine with the JSON/Python dict in the first paragraph of the answer. Try create a view, and a template e.g. titles.html and just pass in the JSON data you have. Post another question if you get stuck I'll help if I see it!

bakkal Over a year ago

If you're thinking about "Model" in MVC pattern, in Django it's not required to be a Django ORM Model, it can be a Python object, Python dict (e.g. JSON). You don't need a Django model to pass data to the template.

Silidrone Over a year ago

This does not answer the question asked, either edit the question's title or edit the answer... It is misleading at best.

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