Creating a JSON response using Django and Python

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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:

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15条回答

我在风中等你

2020-11-22 06:28

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

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臣服心动

2020-11-22 06:28

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')

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心在旅途

2020-11-22 06:32

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'})

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谎友^

2020-11-22 06:36

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)

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误落风尘

2020-11-22 06:36
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" )
```
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逝去的感伤

2020-11-22 06:39

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>

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