Creating a JSON response using Django and Python

Question

I’m trying to convert a server side Ajax response script into a Django HttpResponse, but apparently it’s not working.

This is the server-side script:

/* RECEIVE VALUE */
$validateValue=$_POST['validateValue'];
$validateId=$_POST['validateId'];
$validateError=$_POST['validateError'];

/* RETURN VALUE */
$arrayToJs = array();
$arrayToJs[0] = $validateId;
$arrayToJs[1] = $validateError;

if($validateValue =="Testuser"){  // Validate??
    $arrayToJs[2] = "true";       // RETURN TRUE
    echo '{"jsonValidateReturn":'.json_encode($arrayToJs).'}';  // RETURN ARRAY WITH success
}
else{
    for($x=0;$x<1000000;$x++){
        if($x == 990000){
            $arrayToJs[2] = "false";
            echo '{"jsonValidateReturn":'.json_encode($arrayToJs).'}';   // RETURNS ARRAY WITH ERROR.
        }
    }
}

And this is the converted code

def validate_user(request):
    if request.method == 'POST':
        vld_value = request.POST.get('validateValue')
        vld_id = request.POST.get('validateId')
        vld_error = request.POST.get('validateError')

        array_to_js = [vld_id, vld_error, False]

        if vld_value == "TestUser":
            array_to_js[2] = True
            x = simplejson.dumps(array_to_js)
            return HttpResponse(x)
        else:
            array_to_js[2] = False
            x = simplejson.dumps(array_to_js)
            error = 'Error'
            return render_to_response('index.html',{'error':error},context_instance=RequestContext(request))
    return render_to_response('index.html',context_instance=RequestContext(request))

I’m using simplejson to encode the Python list (so it will return a JSON array). I couldn’t figure out the problem yet. But I think that I did something wrong about the ‘echo’.

Asked By: Switch

||

Source

Answer 1

I usually use a dictionary, not a list to return JSON content.

import json

from django.http import HttpResponse

response_data = {}
response_data['result'] = 'error'
response_data['message'] = 'Some error message'

Pre-Django 1.7 you’d return it like this:

return HttpResponse(json.dumps(response_data), content_type="application/json")

For Django 1.7+, use JsonResponse as shown in this SO answer like so :

from django.http import JsonResponse
return JsonResponse({'foo':'bar'})

Answered By: Tom

Answer 2

I use this, it works fine.

from django.utils import simplejson
from django.http import HttpResponse

def some_view(request):
    to_json = {
        "key1": "value1",
        "key2": "value2"
    }
    return HttpResponse(simplejson.dumps(to_json), mimetype='application/json')

Alternative:

from django.utils import simplejson

class JsonResponse(HttpResponse):
    """
        JSON response
    """
    def __init__(self, content, mimetype='application/json', status=None, content_type=None):
        super(JsonResponse, self).__init__(
            content=simplejson.dumps(content),
            mimetype=mimetype,
            status=status,
            content_type=content_type,
        )

In Django 1.7 JsonResponse objects have been added to the Django framework itself which makes this task even easier:

from django.http import JsonResponse
def some_view(request):
    return JsonResponse({"key": "value"})

Answered By: Dingo

Answer 3

You’ll want to use the django serializer to help with unicode stuff:

from django.core import serializers

json_serializer = serializers.get_serializer("json")()
    response =  json_serializer.serialize(list, ensure_ascii=False, indent=2, use_natural_keys=True)
    return HttpResponse(response, mimetype="application/json")

Answered By: ReduxDJ

Answer 4

How to use google app engine with ajax (json)?

Code Javascript with JQuery:

$.ajax({
    url: '/ajax',
    dataType : 'json',
    cache: false,
    success: function(data) {
        alert('Load was performed.'+data.ajax_resp);
    }
});

Code Python

class Ajax(webapp2.RequestHandler):
    def get(self):
        my_response = {'ajax_resp':'Hello, webapp World!'}
        datos = json.dumps(my_response)

        self.response.headers.add_header('content-type', 'application/json', charset='utf-8')
        self.response.out.write(datos)

Answered By: Samiro

Answer 5

This is my preferred version using a class based view.
Simply subclass the basic View and override the get()-method.

import json

class MyJsonView(View):

    def get(self, *args, **kwargs):
        resp = {'my_key': 'my value',}
        return HttpResponse(json.dumps(resp), mimetype="application/json" )

Answered By: droidballoon

Answer 6

from django.http import HttpResponse
import json

class JsonResponse(HttpResponse):
    def __init__(self, content={}, mimetype=None, status=None,
             content_type='application/json'):
        super(JsonResponse, self).__init__(json.dumps(content), mimetype=mimetype,
                                           status=status, content_type=content_type)

And in the view:

resp_data = {'my_key': 'my value',}
return JsonResponse(resp_data)

Answered By: Dmitry Demidenko

Answer 7

New in django 1.7

you could use JsonResponse objects.

from the docs:

from django.http import JsonResponse
return JsonResponse({'foo':'bar'})

Answered By: srj

Answer 8

Since Django 1.7 you have a standard JsonResponse that’s exactly what you need:

from django.http import JsonResponse
...
return JsonResponse(array_to_js, safe=False)

You don’t even need to json.dump your array.

Answered By: Akhorus

Answer 9

Django code views.py:

def view(request):
    if request.method == 'POST':
        print request.body
        data = request.body
        return HttpResponse(json.dumps(data))

HTML code view.html:

<!DOCTYPE html>
<html>
<head>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<script>
$(document).ready(function(){
    $("#mySelect").change(function(){
        selected = $("#mySelect option:selected").text()
        $.ajax({
            type: 'POST',
            dataType: 'json',
            contentType: 'application/json; charset=utf-8',
            url: '/view/',
            data: {
                    'fruit': selected
                  },
            success: function(result) {
                        document.write(result)
                    }
    });
  });
});
</script>
</head>
<body>

<form>
    {{data}}
    <br>
Select your favorite fruit:
<select id="mySelect">
  <option value="apple" selected >Select fruit</option>
  <option value="apple">Apple</option>
  <option value="orange">Orange</option>
  <option value="pineapple">Pineapple</option>
  <option value="banana">Banana</option>
</select>
</form>
</body>
</html>

Answered By: Rajan Mandanka

Answer 10

For those who use Django 1.7+

from django.http import JsonResponse

def your_view(request):
    json_object = {'key': "value"}
    return JsonResponse(json_object)

official docs

Answered By: Andres

Answer 11

In View use this:

form.field.errors|striptags

for getting validation messages without html

Answered By: Deepak Sharma

Answer 12

Its very convenient with Django version 1.7 or higher as you have the JsonResponse class, which is a subclass of HttpResponse.

from django.http import JsonResponse
    def profile(request):
        data = {
            'name': 'Raghav',
            'location': 'India',
            'is_active': False,
            'count': 28
        }
        return JsonResponse(data)

For older versions of Django, you must use an HttpResponse object.

import json
from django.http import HttpResponse

def profile(request):
    data = {
        'name': 'Raghav',
        'location': 'India',
        'is_active': False,
        'count': 28
    }
    dump = json.dumps(data)
    return HttpResponse(dump, content_type='application/json')

Answered By: Tanmay D

Answer 13

With Django Class-based views you can write:

from django.views import View
from django.http import JsonResponse

class JsonView(View):
    def get(self, request):
        return JsonResponse({'some': 'data'})

and with Django-Rest-Framework you can write:

from rest_framework.views import APIView
from rest_framework.response import Response

class JsonView(APIView):
    def get(self, request):
        return Response({'some': 'data'})

Answered By: elim

Answer 14

First import this:

from django.http import HttpResponse

If you have the JSON already:

def your_method(request):
    your_json = [{'key1': value, 'key2': value}]
    return HttpResponse(your_json, 'application/json')

If you get the JSON from another HTTP request:

def your_method(request):
    response = request.get('https://www.example.com/get/json')
    return HttpResponse(response, 'application/json')

Answered By: Sakthivel Karthikeyan

Answer 15

Most of these answers are out of date. JsonResponse is not recommended because it escapes the characters, which is usually undesired. Here’s what I use:

views.py (returns HTML)

from django.shortcuts import render
from django.core import serializers

def your_view(request):
    data = serializers.serialize('json', YourModel.objects.all())
    context = {"data":data}
    return render(request, "your_view.html", context)

views.py (returns JSON)

from django.core import serializers
from django.http import HttpResponse

def your_view(request):
    data = serializers.serialize('json', YourModel.objects.all())
    return HttpResponse(data, content_type='application/json')

Bonus for Vue Users

If you want to bring your Django Queryset into Vue, you can do the following.

template.html

<div id="dataJson" style="display:none">
{{ data }}
</div>

<script>
let dataParsed = JSON.parse(document.getElementById('dataJson').textContent);
var app = new Vue({
  el: '#app',
  data: {
    yourVariable: dataParsed,
  },
})
</script>

Answered By: devdrc

Answer 16

def your_view(request):
    response = {'key': "value"}
    return JsonResponse(json.dumps(response), content_type="application/json",safe=False)

#Specify the content_type and use json.dump() son as the content not to be sent as object

Answered By: NINSIIMA WILBER

Answer 17

This way the json contents can be downloaded as a file with a specific filename.

import json
from django.http import HttpResponse

def download_json(request):
    data = {'some': 'information'}

    # serialize data obj as a JSON stream 
    data = json.dumps(data)
    response = HttpResponse(data, content_type='application/json charset=utf-8')

    # add filename to response
    response['Content-Disposition'] = 'attachment; filename="filename.json"'
    return response

Answered By: Joe Web

Creating a JSON response using Django and Python

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