I have the following DJango view
def company(request):
company_list = Company.objects.all()
output = serializers.serialize('json', company_list, fields=('name','phonenumber','email','companylogo'))
return HttpResponse(output, content_type="application/json")
it result as follows:
[{"pk": 1, "model": "test.company", "fields": {"companylogo": null, "phonenumber": "741.999.5554", "name": "Remax", "email": "home#remax.co.il"}}, {"pk": 4, "model": "test.company", "fields": {"companylogo": null, "phonenumber": "641-7778889", "name": "remixa", "email": "a#aa.com"}}, {"pk": 2, "model": "test.company", "fields": {"companylogo": null, "phonenumber": "658-2233444", "name": "remix", "email": "b#bbb.com"}}, {"pk": 7, "model": "test.company", "fields": {"companylogo": null, "phonenumber": "996-7778880", "name": "remix", "email": "a#aba.com"}}]
my questions:
1. can i control the order of the fields
2. can i change the name of the fields
3. I was expecting to see the result with indentation in the browser i.e. instead of one long line to see something like this:
[
{
"pk": 1,
"model": "test.company",
"fields":
{
"companylogo": null,
"phonenumber": "741.999.5554",
"name": "Remax",
"email": "home#remax.co.il"
}
},
{
"pk": 4,
"model": "test.company",
"fields":
{
"companylogo": null,
"phonenumber": "641-7778889",
"name": "remixa",
"email": "a#aa.com"
}
},
....
}
]
you can get pretty format in this way:
return JsonResponse(your_json_dict, json_dumps_params={'indent': 2})
django doc as the first comment say
Python (unrelated to Django and starting with 2.6) has a built in json library that can accomplish the indentation you require. If you are looking for something quick and dirty for debug purposes you could do something like this:
from django.http import HttpResponse
from django.core import serializers
import json
def company(request, pretty=False):
company_list = Company.objects.all()
output = serializers.serialize('json', company_list, fields=('name','phonenumber','email','companylogo'))
if pretty:
output = json.dumps(json.loads(output), indent=4))
return HttpResponse(output, content_type="application/json")
But this is a performance issue if the Company model is large. I recommend taking Dan R's advice and use a browser plugin to parse and render the json or come up with some other client side solution. I have a script that takes in a json file and does exactly the same thing as the code above, reads in the json and prints it out with indent=4.
As for sorting your output, you can just use the order_by method on your query set:
def company(request):
company_list = Company.objects.order_by("sorting_field")
...
And if you always want that model sorted that way, you can use the ordering meta-class option:
class Company(models.Model):
class Meta:
ordering = ["sorting_field"]
...
As a final note, If your intent is to expose your models with a web service, I highly recommend taking a look at tastypie. It may help you in the long run as it provides many other convenient features that help towards that end.
With Django 1.7 I can get nicely indented JSON by using the indent parameter of the serializer. For instance, in a command that dumps data from my database:
self.stdout.write(serializers.serialize("json",
records,
indent=2))
The indent parameter has been in Django since version 1.5. The output I get looks like this:
[
{
"fields": {
"type": "something",
"word": "something else",
},
"model": "whatever",
"pk": 887060
},
{
"fields": {
"type": "something more",
"word": "and another thing",
},
"model": "whatever",
"pk": 887061
},
...
To order your records then you'd have to do what Kevin suggested and use order_by, or whatever method you want to order the records you pass to the serializer. For instance, I use itertools.chain to order different querysets that return instances of different models.
The serializer does not support ordering fields according to your preferences, or renaming them. You have to write your own code to do this or use an external tool.
JSON doesn't have indentation, it's simply structured data. Browsers or other tools may format the JSON so that it looks nice but by default it's not there. It's also not part of the JSON as the formatting is just how it looks on the screen. JSON is often processed by other code or services so they don't care about indentation, as long as the data is structured correctly.
Related
I have 2 APIs from my existing project. Where One provides the latest blog posts and another one provides sorting details. The 2nd API (sorting) gives blog posts ID and an ordering number, which should be in the 1st,2nd,3rd...n position. If I filter in the first API with that given ID I can get the blog post details.
How can I create a Django REST API from Database? or an API merging from that 2 APIs? Any tutorial or reference which might help me?
Frist API Response:
{
"count": 74,
"next": "https://cms.example.com/api/v2/stories/?page=2",
"previous": null,
"results": [
{
"id": 111,
"meta": {
"type": "blog.CreateStory",
"seo_title": "",
"search_description": "",
"first_published_at": "2022-10-09T07:29:17.029746Z"
},
"title": "A Test Blog Post"
},
{
"id": 105,
"meta": {
"type": "blog.CreateStory",
"seo_title": "",
"search_description": "",
"first_published_at": "2022-10-08T04:45:32.165072Z"
},
"title": "Blog Story 2"
},
2nd API Response
[
{
"featured_item": 1,
"sort_order": 0,
"featured_page": 105
},
{
"featured_item": 1,
"sort_order": 1,
"featured_page": 90
},
Here I want to create another API that will provide more details about sorting for example it will sort like this https://cms.example.com/api/v2/stories/105 and catch Title, Image & Excerpt and If there is no data from Sorting details it will show the first API's response by default.
After searching, I found that you can make API from Database. In setting you need to set the database credentials and then need to create a class inside your models.py and inside class's meta you need to set meta name to db_table and then create serializers.py and views.py as you create REST API.
class SortAPI(models.Model):
featured_item_id = models.IntegerField()
sort_order = models.IntegerField()
title=models.TextField()
first_published_at=models.DateTimeField()
alternative_title= models.TextField()
excerpt=models.TextField()
sub_heading=models.TextField()
news_slug=models.TextField()
img_title=models.TextField()
img_url=models.TextField()
img_width=models.IntegerField()
img_height=models.IntegerField()
class Meta:
db_table = 'view_featured'
I have a simple model like:
class MyModel(models.Model):
data = JSONField()
The JSONField data is in the following structure:
{
"name": "Brian",
"skills": [
{"id": 4, "name": "First aid"},
{"id": 5, "name": "Second aid"}
]
}
I'd like to create a query that gets a list of MyModels filtered by the id of the skill inside the data.
I've tried a few different avenues here, and can do the work in Python but I'm pretty sure there's a way to do this in Django; I think my SQL isn't good enough to figure it out.
Cheers in advance.
try this
>>> MyModel.objects.filter(data__skills__contains=[{'id':4}, {'id':5}])
more about JSON filter https://docs.djangoproject.com/en/3.1/topics/db/queries/#querying-jsonfield
I want to import data that is in regular JSON format into a (SQLite) db within Django. It looks like fixtures is the recommended way of doing that. I'm not very clear on the steps I need take. Specifically, do I need to create a mapping of which field should go into which model (assuming my model is already defined in models.py) etc? The Django example looks like this:
[
{
"model": "myapp.person",
"pk": 1,
"fields": {
"first_name": "John",
"last_name": "Lennon"
} }, {
"model": "myapp.person",
"pk": 2,
"fields": {
"first_name": "Paul",
"last_name": "McCartney"
}
} ]
However my data isn't in this format. Do I need to edit the JSON file to look like this? Should I just abort and import json instead?
json.loads is all that would be needed. Something like
as_object_list = json.loads(your_json)
This converts the json to a dict, from there you can can create the model by doing
YourModel.objects.create(**as_object_list)
It seems to me that there should be an automatic way to query the results of a Django Rest Framework call and operate it like a dictionary (or something similar). Am I missing something, or is that not possible?
i.e.,
if the call to http://localhost:8000/api/1/roles/
yields
{"count": 2, "next": null, "previous": null, "results": [{"user": {"username": "smithb", "first_name": "Bob", "last_name": "Smith"}, "role_type": 2, "item": 1}, {"user": {"username": "jjones", "first_name": "Jane", "last_name": "Jones"}, "role_type": 2, "item": 1}]}
I would think something akin to http://localhost:8000/api/1/roles/0/user/username should return smithb.
Does this functionality exist or do I need to build it myself?
It appears to be something you will have to build yourself. That said Django makes this kind of thing very easy. In URLS you can specify parts of the url path to pass to the view. You can catch the values using regex and then pass them into your views function.
Urls:
url(regex=r'^user/api/1/roles/(?P<usernumber>\w{1,50})/(?P<username>\w{1,50})/$', view='views.profile_page')
a request for http://domain/user/api/1/roles/0/username/
View:
def someApiFunction(request, usernumber=None ,username=None):
return HttpResponse(username)
Some additional Resources:
https://docs.djangoproject.com/en/1.7/intro/tutorial03/#writing-more-views
Capturing url parameters in request.GET
so I just try to use tastypie put method to edit objects.
let's say my object have this structure:
{
"id": 38,
"media": [],
"name": "tesdr",
"resource_uri": "/api/v2/group/38/",
"status": 7,
"user_name": null,
"users": []
}
witch media and users are related many to many field. when I edit group and use put without any change in m2m fields every thing works fine.
but when I try to put something like this:
{
"id": 38,
"media": [
"/api/v2/media/70/"
],
"name": "testgpat",
"resource_uri": "/api/v2/group/40/",
"status": 6,
"user_name": null,
"users": []
}
tastypie return an 401 http error. so what is the solution? any idea?
ok! I just solved the problem. have to define a many to many field in both resources which wants to set relation.
thanks all! :D