CODEWITHSUNDEEP
djangosimplejson
iJK
iJK

Reputation: 4765

Removing fields from json

I have a view function which renders json. I am able to specify which columns I want in my json but I don't know how to change the name of the key fields. Like the field "pk" should be "id".

I am using this autocomplete control (http://loopj.com/2009/04/25/jquery-plugin-tokenizing-autocomplete-text-entry/) and it requires the json to have certain fields.

from django.http import HttpResponse
from django.shortcuts import render_to_response
from iCookItThisWay.recipes import models
from django.core import serializers
from django.utils import simplejson

def index(request, template_name):
    meal_types = []
    q = ''

    if 'q' in request.GET and request.GET['q']:
        q = request.GET['q']

    if len(q) > 0:
        meal_types = models.MealType.objects.filter(name__istartswith=q)

    json_serializer = serializers.get_serializer("json")()
    sdata = json_serializer.serialize(meal_types, ensure_ascii=False, fields = ('id', 'name'))

    return HttpResponse(simplejson.dumps(sdata), mimetype='application/json')

Could you also please point me to some documentation. I feel that I am crap at finding documentation.

Upvotes: 3

Views: 1348

Answers (1)

Chris Lawlor
Chris Lawlor

Reputation: 48922

Instead of using the serializer, you can build a dict manually and convert it to json via .dumps()

meal_types = models.MealType.objects.filter(name__istartswith=q)
results = []
for meal_type in meal_types:
    results.append(
        {'id': meal_type.id,
         'name': meal_type.name})

return HttpResponse(simplejson.dumps(results), mimetype='application/json')

You could also build the results with a list comprehension, since there are only a couple of fields:

results = [{'id': mt.id, 'name': mt.name} for mt in meal_types]

Upvotes: 1

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