I need to convert a Django Queryset Object into a Json string. The built in Django Serialization library works great. Although it specifies the name of the Model from where it was created. Since I don't need this, how do I get rid of it? What else do I need to override to be able to use the overridden end_object method below?
class Serializer(PythonSerializer):
def end_object(self, obj):
self.objects.append({
"model" : smart_unicode(obj._meta), # <-- I want to remove this
"pk" : smart_unicode(obj._get_pk_val(), strings_only=True),
"fields" : fields
})
self._current = None
Sorry I had totally forgot about this question. This is how I ended up solving it (with thanks to FunkyBob on #django):
from django.core.serializers.python import Serializer
class MySerialiser(Serializer):
def end_object( self, obj ):
self._current['id'] = obj._get_pk_val()
self.objects.append( self._current )
# views.py
serializer = MySerialiser()
data = serializer.serialize(some_qs)
Here's a serializer that removes all metadata (pk, models) from the serialized output, and moves the fields up to the top-level of each object. Tested in django 1.5
from django.core.serializers import json
class CleanSerializer(json.Serializer):
def get_dump_object(self, obj):
return self._current
Serializer = CleanSerializer
Edit:
In migrating my app to an older version of django (precipitated by a change in hosting provider), the above serializer stopped working. Below is a serializer that handles the referenced versions:
import logging
import django
from django.core.serializers import json
from django.utils.encoding import smart_unicode
class CleanSerializer148(json.Serializer):
def end_object(self, obj):
current = self._current
current.update({'pk': smart_unicode(obj._get_pk_val(),
strings_only=True)})
self.objects.append(current)
self._current = None
class CleanSerializer151(json.Serializer):
def get_dump_object(self, obj):
self._current['pk'] = obj.pk
return self._current
if django.get_version() == '1.4.8':
CleanSerializer = CleanSerializer148
else:
CleanSerializer = CleanSerializer151
Serializer = CleanSerializer
Override JSON serializer class:
from django.core.serializers.json import Serializer, DjangoJSONEncoder
from django.utils import simplejson
class MySerializer(Serializer):
"""
Convert QuerySets to JSONS, overrided to remove "model" from JSON
"""
def end_serialization(self):
# little hack
cleaned_objects = []
for obj in self.objects:
del obj['model']
cleaned_objects.append(obj)
simplejson.dump(cleaned_objects, self.stream, cls=DjangoJSONEncoder, **self.options)
In the view:
JSONSerializer = MySerializer
jS = JSONSerializer()
Related
I am working on Django React Project using the Django REST FRAMEWORK,I am trying to post some data tied to my model.
The list view and the detail view of the project works pretty fine,The only problem is when I try to make a POST request.
Whenever I Try post the data in the CreateAPIView I get an error :
Got a `TypeError` when calling `Article.objects.create()`. This may be because
you have a writable field on the serializer class that is not a valid argument to
`Article.objects.create()`. You may need to make the field read-only, or override
the ArticleSerializer.create() method to handle this correctly.
I have searched through various past problems but non of them seem to fix my problem.
Here is my serializers file:
from rest_framework import serializers
from articles.models import Article
class ArticleSerializer(serializers.ModelSerializer):
class Meta:
model = Article
fields = ('id','title','content','star_count','like_count','comment_count','avatar')
Here is my views file
from rest_framework.generics import ListAPIView,RetrieveAPIView,CreateAPIView,UpdateAPIView,DestroyAPIView
from .serializers import ArticleSerializer
from articles.models import Article
from rest_framework import viewsets
class ArticleViewSets(viewsets.ModelViewSet):
queryset = Article.objects.all()
serializer_class = ArticleSerializer
models file
content = models.TextField()
comment_count = models.IntegerField(default=0,null=True,blank=True)
like_count = models.IntegerField(default=0,null=True,blank=True)
star_count = models.IntegerField(default=0,null=True,blank=True)
avatar = models.ImageField(null=True,blank=True)
def __str__(self):
return self.title
def save(self):
if not self.slug:
self.slug = slugify(self.title)
super(Article,self).save()
Here is the error generated when I try to make a POST request based on the django rest framework createAPIVew
Got a `TypeError` when calling `Article.objects.create()`. This may be because you have a writable field on the serializer class that is not a valid argument to `Article.objects.create()`. You may need to make the field read-only, or override the ArticleSerializer.create() method to handle this correctly.
I have Django API built in and I have endpoint the return all object. I want the user to provide me with keyword to filter this queryset. What is the best way to do it. and how to do it plz ?
is it in get_queryset? if yes can you help me !?
You have access to the GET parameters (in the querystring) with self.request.GET [Django-doc].
So for example if there is a parameter ?category=foo, you can access foo with self.request.GET['category'], or self.request.GET.get('category') if you want it to return None in case it is missing.
You thus can filter for example with:
from rest_framework import generics
from app.models import SomeModel
from app.serializers import SomeSerializer
class UserList(generics.ListAPIView):
model = SomeModel
def get_queryset(self):
qs = super().get_queryset()
category = self.request.GET.get('category')
if category is None:
return qs
return qs.filter(category=categry)
just give pass the parameter with some default value:
def get_queryset(self, some_thing=default):
.
.
.
It will work
I have a very basic Django Rest API.
I don't know how to have some HTML views, in the same django project, which uses API (finally keeping API returning JSON only).
I followed this, but it seems to change the API View (in this case, curl will retrieve HTML and not JSON) :
https://www.django-rest-framework.org/api-guide/renderers/#templatehtmlrenderer
Do I need another Django App ? Another Django project ? Some JS ?
EDIT :
Ok, I've seen it's possible, thanks to rrebase.
But I can't retrieve the JSON with Curl, here my views.py
from rest_framework import generics
from rest_framework.renderers import TemplateHTMLRenderer, JSONRenderer
from rest_framework.response import Response
from rest_framework.views import APIView
from rest_framework.permissions import IsAdminUser
from . import models
from . import serializers
class UserListView(generics.ListAPIView):
renderer_classes = [JSONRenderer, TemplateHTMLRenderer]
template_name = 'profile_list.html'
def get(self, request):
queryset = models.CustomUser.objects.all()
serializer_class = serializers.UserSerializer
return Response({'profiles': queryset})
My models.py
from django.db import models
from django.contrib.auth.models import AbstractUser
class CustomUser(AbstractUser):
def __str__(self):
return self.email
I get an error "Object of type 'CustomUser' is not JSON serializable" when I request the API (http://127.0.0.1:8000/api/v1/users/)
Sorry, it's some different that initial question...
Yes, you can have both. The link you provided to docs has the following:
You can use TemplateHTMLRenderer either to return regular HTML pages using REST framework, or to return both HTML and API responses from a single endpoint.
When making an API request, set the ACCEPT request-header accordingly to html or json.
Finally I made some conditions in my view, and it's working
class UserListView(generics.ListAPIView):
renderer_classes = [JSONRenderer, TemplateHTMLRenderer]
permission_classes = (IsAdminUser,)
def get(self, request):
queryset = CustomUser.objects.all()
if request.accepted_renderer.format == 'html':
data = {'profiles': queryset}
return Response(data, template_name='profile_list.html')
else:
serializer = UserSerializer(queryset, many=True)
data = serializer.data
return Response(data)
This is an offshoot of the generic how can I include properties in a JSON serialization, which is answered here: https://stackoverflow.com/a/38253327/4140357
from django.core.serializers.base import Serializer as BaseSerializer
from django.core.serializers.python import Serializer as PythonSerializer
from django.core.serializers.json import Serializer as JsonSerializer
class ExtBaseSerializer(BaseSerializer):
def serialize_property(self, obj):
model = type(obj)
for field in self.selected_fields:
if hasattr(model, field) and type(getattr(model,
field)) == property:
self.handle_prop(obj, field)
def handle_prop(self, obj, field):
self._current[field] = getattr(obj, field)
def end_object(self, obj):
self.serialize_property(obj)
super(ExtBaseSerializer, self).end_object(obj)
class ExtPythonSerializer(ExtBaseSerializer, PythonSerializer):
pass
class ExtJsonSerializer(ExtPythonSerializer, JsonSerializer):
pass
How to use it:
ExtJsonSerializer().serialize(
MyModel.objects.all(),
fields=['field_name_1', 'property_1' ...]
)
And it works great for JSON.
How can you do the same thing for GEODJango's GEJSON serializer?
It is quite straightforward to alter the original answer for JSON to work with GEOJSON.
You need to change one line in the imports so instead of importing
from django.core.serializers.json import Serializer as JsonSerializer you import the GEOJSON one
from django.contrib.gis.serializers.geojson import Serializer as JsonSerializer
Here's the full code for ease of cut-n-paste.
from django.core.serializers.base import Serializer as BaseSerializer
from django.core.serializers.python import Serializer as PythonSerializer
from django.contrib.gis.serializers.geojson import Serializer as JsonSerializer
class ExtBaseSerializer(BaseSerializer):
def serialize_property(self, obj):
model = type(obj)
for field in self.selected_fields:
if hasattr(model, field) and type(getattr(model,
field)) == property:
self.handle_prop(obj, field)
def handle_prop(self, obj, field):
self._current[field] = getattr(obj, field)
def end_object(self, obj):
self.serialize_property(obj)
super(ExtBaseSerializer, self).end_object(obj)
class ExtPythonSerializer(ExtBaseSerializer, PythonSerializer):
pass
class ExtJsonSerializer(ExtPythonSerializer, JsonSerializer):
pass
How to use it:
ExtJsonSerializer().serialize(
MyModel.objects.all(),
fields=['field_name_1', 'property_1' ...]
)
My app has users who create pages. In the Page screen of the admin, I'd like to list the User who created the page, and in that list, I'd like the username to have a link that goes to the user page in admin (not the Page).
class PageAdmin(admin.ModelAdmin):
list_display = ('name', 'user', )
list_display_links = ('name','user',)
admin.site.register(Page, PageAdmin)
I was hoping that by making it a link in the list_display it would default to link to the actual user object, but it still goes to Page.
I'm sure I'm missing something simple here.
Modifying your model isn't necessary, and it's actually a bad practice (adding admin-specific view-logic into your models? Yuck!) It may not even be possible in some scenarios.
Luckily, it can all be achieved from the ModelAdmin class:
from django.urls import reverse
from django.utils.safestring import mark_safe
class PageAdmin(admin.ModelAdmin):
# Add it to the list view:
list_display = ('name', 'user_link', )
# Add it to the details view:
readonly_fields = ('user_link',)
def user_link(self, obj):
return mark_safe('{}'.format(
reverse("admin:auth_user_change", args=(obj.user.pk,)),
obj.user.email
))
user_link.short_description = 'user'
admin.site.register(Page, PageAdmin)
Edit 2016-01-17:
Updated answer to use make_safe, since allow_tags is now deprecated.
Edit 2019-06-14:
Updated answer to use django.urls, since as of Django 1.10 django.core.urls has been deprecated.
Add this to your model:
def user_link(self):
return '%s' % (reverse("admin:auth_user_change", args=(self.user.id,)) , escape(self.user))
user_link.allow_tags = True
user_link.short_description = "User"
You might also need to add the following to the top of models.py:
from django.template.defaultfilters import escape
from django.core.urls import reverse
In admin.py, in list_display, add user_link:
list_display = ('name', 'user_link', )
No need for list_display_links.
You need to use format_html for modern versions of django
#admin.register(models.Foo)
class FooAdmin(admin.ModelAdmin):
list_display = ('ts', 'bar_link',)
def bar_link(self, item):
from django.shortcuts import resolve_url
from django.contrib.admin.templatetags.admin_urls import admin_urlname
url = resolve_url(admin_urlname(models.Bar._meta, 'change'), item.bar.id)
return format_html(
'{name}'.format(url=url, name=str(item.bar))
)
I ended up with a simple helper:
from django.shortcuts import resolve_url
from django.utils.safestring import SafeText
from django.contrib.admin.templatetags.admin_urls import admin_urlname
from django.utils.html import format_html
def model_admin_url(obj: Model, name: str = None) -> str:
url = resolve_url(admin_urlname(obj._meta, SafeText("change")), obj.pk)
return format_html('{}', url, name or str(obj))
Then you can use the helper in your model-admin:
class MyAdmin(admin.ModelAdmin):
readonly_field = ["my_link"]
def my_link(self, obj):
return model_admin_url(obj.my_foreign_key)
I needed this for a lot of my admin pages, so I created a mixin for it that handles different use cases:
pip install django-admin-relation-links
Then:
from django.contrib import admin
from django_admin_relation_links import AdminChangeLinksMixin
#admin.register(Group)
class MyModelAdmin(AdminChangeLinksMixin, admin.ModelAdmin):
# ...
change_links = ['field_name']
See the GitHub page for more info. Try it out and let me know how it works out!
https://github.com/gitaarik/django-admin-relation-links
I decided to make a simple admin mixin that looks like this (see docstring for usage):
from django.contrib.contenttypes.models import ContentType
from django.utils.html import format_html
from rest_framework.reverse import reverse
class RelatedObjectLinkMixin(object):
"""
Generate links to related links. Add this mixin to a Django admin model. Add a 'link_fields' attribute to the admin
containing a list of related model fields and then add the attribute name with a '_link' suffix to the
list_display attribute. For Example a Student model with a 'teacher' attribute would have an Admin class like this:
class StudentAdmin(RelatedObjectLinkMixin, ...):
link_fields = ['teacher']
list_display = [
...
'teacher_link'
...
]
"""
link_fields = []
def __init__(self, *args, **kwargs):
super().__init__(*args, **kwargs)
if self.link_fields:
for field_name in self.link_fields:
func_name = field_name + '_link'
setattr(self, func_name, self._generate_link_func(field_name))
def _generate_link_func(self, field_name):
def _func(obj, *args, **kwargs):
related_obj = getattr(obj, field_name)
if related_obj:
content_type = ContentType.objects.get_for_model(related_obj.__class__)
url_name = 'admin:%s_%s_change' % (content_type.app_label, content_type.model)
url = reverse(url_name, args=[related_obj.pk])
return format_html('{}', url, str(related_obj))
else:
return None
return _func
If anyone is trying to do this with inline admin, consider a property called show_change_link since Django 1.8.
Your code could then look like this:
class QuestionInline(admin.TabularInline):
model = Question
extra = 1
show_change_link = True
class TestAdmin(admin.ModelAdmin):
inlines = (QuestionInline,)
admin.site.register(Test, TestAdmin)
This will add a change/update link for each foreign key relationship in the admin's inline section.