I'm trying to add custom functionality to django router methods.
This is my router that exposes the standard methods on an user.
class UserViewSet(viewsets.ModelViewSet):
queryset = User.objects.all()
serializer_class = UserSerializer
permission_classes = [BasePermission]
I'm validating the user using serializer validation methods.
class UserSerializer(serializers.ModelSerializer):
password = serializers.CharField(write_only=True)
MOBILE_ERROR = 'Mobile number should be 10 digits long and only contain numbers.'
EMAIL_ERROR = 'Incorrect email format'
USERNAME_ERROR = 'Username must be at least 6 characters long and contain only letters and numbers.'
class Meta:
model = User
fields = '__all__'
def validate_mobile(self, value):
regexp = re.compile(r'^[0-9]{10}$')
if regexp.search(value):
return value
raise serializers.ValidationError(self.MOBILE_ERROR)
def validate_email(self, value):
if validate_email(value):
return value
raise serializers.ValidationError(self.EMAIL_ERROR)
def validate_username(self, value):
regexp = re.compile(r'^[a-zA-Z0-9]{6,}$')
if regexp.search(value):
return value
raise serializers.ValidationError(self.USERNAME_ERROR)
And this is my route.
router = DefaultRouter(trailing_slash=False)
router.register(r'user', UserViewSet),
urlpatterns = router.urls
I want to add a method send_activation_code if the user is created successfully. How do I do this?
For such purpose you can use signals. Every time when your app creates new User instance - some action should be performed. In your case you should connect build-in signal post_save and your existed send_activation_code function
Example for your case:
yourapp/signals.py:
from django.contrib.auth.models import User
from django.db.models.signals import post_save
from django.dispatch import receiver
#receiver(post_save, sender=User)
def send_activation_code_signal(sender, instance, created, **kwargs):
if created:
send_activation_code(instance.phone_number)
Also, you need to import signals in your app config file
yourapp/app.py:
from django.apps import AppConfig
from django.utils.translation import ugettext_lazy as _
class YourAppConfig(AppConfig):
name = 'yourproject.yourapp'
verbose_name = _('yourapp')
def ready(self):
import yourproject.yourapp.signals
yourapp/__init__.py:
default_app_config = 'yourproject.yourapp.apps.YourAppConfig'
If you dont need to send code every time User instance created - you can specify more statements, for example:
if created and instance.validated:
send_activation_code(instance.phone_number)
There are some more useful built-in signals in Django, check docs
Django signals docs: https://docs.djangoproject.com/en/3.0/ref/signals/
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 done the below post_save signal in my project.
from django.db.models.signals import post_save
from django.contrib.auth.models import User
# CORE - SIGNALS
# Core Signals will operate based on post
def after_save_handler_attr_audit_obj(sender, **kwargs):
print User.get_profile()
if hasattr(kwargs['instance'], 'audit_obj'):
if kwargs['created']:
kwargs['instance'].audit_obj.create(operation="INSERT", operation_by=**USER.ID**).save()
else:
kwargs['instance'].audit_obj.create(operation="UPDATE").save()
# Connect the handler with the post save signal - Django 1.2
post_save.connect(after_save_handler_attr_audit_obj, dispatch_uid="core.models.audit.new")
The operation_by column, I want to get the user_id and store it. Any idea how can do that?
Can't be done. The current user is only available via the request, which is not available when using purely model functionality. Access the user in the view somehow.
I was able to do it by inspecting the stack and looking for the view then looking at the local variables for the view to get the request. It feels like a bit of a hack, but it worked.
import inspect, os
#receiver(post_save, sender=MyModel)
def get_user_in_signal(sender, **kwargs):
for entry in reversed(inspect.stack()):
if os.path.dirname(__file__) + '/views.py' == entry[1]:
try:
user = entry[0].f_locals['request'].user
except:
user = None
break
if user:
# do stuff with the user variable
Ignacio is right. Django's model signals are intended to notify other system components about events associated with instances and their respected data, so I guess it's valid that you cannot, say, access request data from a model post_save signal, unless that request data was stored on or associated with the instance.
I guess there are lots of ways to handle it, ranging from worse to better, but I'd say this is a prime example for creating class-based/function-based generic views that will automatically handle this for you.
Have your views that inherit from CreateView, UpdateView or DeleteView additionally inherit from your AuditMixin class if they handle verbs that operate on models that need to be audited. The AuditMixin can then hook into the views that successfully create\update\delete objects and create an entry in the database.
Makes perfect sense, very clean, easily pluggable and gives birth to happy ponies. Flipside? You'll either have to be on the soon-to-be-released Django 1.3 release or you'll have to spend some time fiddlebending the function-based generic views and providing new ones for each auditing operation.
You can do that with the help of middleware. Create get_request.py in your app. Then
from threading import current_thread
from django.utils.deprecation import MiddlewareMixin
_requests = {}
def current_request():
return _requests.get(current_thread().ident, None)
class RequestMiddleware(MiddlewareMixin):
def process_request(self, request):
_requests[current_thread().ident] = request
def process_response(self, request, response):
# when response is ready, request should be flushed
_requests.pop(current_thread().ident, None)
return response
def process_exception(self, request, exception):
# if an exception has happened, request should be flushed too
_requests.pop(current_thread().ident, None)
Then add this middleware to your settings:
MIDDLEWARE = [
....
'<your_app>.get_request.RequestMiddleware',
]
Then add import to your signals:
from django.db.models.signals import post_save
from django.contrib.auth.models import User
from <your_app>.get_request import current_request
# CORE - SIGNALS
# Core Signals will operate based on post
def after_save_handler_attr_audit_obj(sender, **kwargs):
print(Current User, current_request().user)
print User.get_profile()
if hasattr(kwargs['instance'], 'audit_obj'):
if kwargs['created']:
kwargs['instance'].audit_obj.create(operation="INSERT", operation_by=**USER.ID**).save()
else:
kwargs['instance'].audit_obj.create(operation="UPDATE").save()
# Connect the handler with the post save signal - Django 1.2
post_save.connect(after_save_handler_attr_audit_obj, dispatch_uid="core.models.audit.new")
Why not adding a middleware with something like this :
class RequestMiddleware(object):
thread_local = threading.local()
def process_request(self, request):
RequestMiddleware.thread_local.current_user = request.user
and later in your code (specially in a signal in that topic) :
thread_local = RequestMiddleware.thread_local
if hasattr(thread_local, 'current_user'):
user = thread_local.current_user
else:
user = None
For traceability add two attributes to your Model(created_by and updated_by), in "updated_by" save the last user who modified the record. Then in your signal you have the user:
models.py:
class Question(models.Model):
question_text = models.CharField(max_length=200)
pub_date = models.DateTimeField('date published')
created_by = models. (max_length=100)
updated_by = models. (max_length=100)
views.py
p = Question.objects.get(pk=1)
p.question_text = 'some new text'
p.updated_by = request.user
p.save()
signals.py
#receiver(pre_save, sender=Question)
def do_something(sender, instance, **kwargs):
try:
obj = Question.objects.get(pk=instance.pk)
except sender.DoesNotExist:
pass
else:
if not obj.user == instance.user: # Field has changed
# do something
print('change: user, old=%s new=%s' % (obj.user, instance.user))
You could also use django-reversion for this purpose, e.g.
from reversion.signals import post_revision_commit
import reversion
#receiver(post_save)
def post_revision_commit(sender, **kwargs):
if reversion.is_active():
print(reversion.get_user())
Read more on their API https://django-reversion.readthedocs.io/en/stable/api.html#revision-api
You can do a small hack by overriding you model save() method and setting the user on the saved instance as additional parameter. To get the user I used get_current_authenticated_user() from django_currentuser.middleware.ThreadLocalUserMiddleware (see https://pypi.org/project/django-currentuser/).
In your models.py:
from django_currentuser.middleware import get_current_authenticated_user
class YourModel(models.Model):
...
...
def save(self, *args, **kwargs):
# Hack to pass the user to post save signal.
self.current_authenticated_user = get_current_authenticated_user()
super(YourModel, self).save(*args, **kwargs)
In your signals.py:
#receiver(post_save, sender=YourModel)
def your_model_saved(sender, instance, **kwargs):
user = getattr(instance, 'current_authenticated_user', None)
PS: Don't forget to add 'django_currentuser.middleware.ThreadLocalUserMiddleware' to your MIDDLEWARE_CLASSES.
I imagine you would have figured this out, but I had the same problem and I realised that all the instances I create had a reference to the user that creates them (which is what you are looking for)
it's possible i guess.
in models.py
class _M(models.Model):
user = models.ForeignKey(...)
in views.py
def _f(request):
_M.objects.create(user=request.user)
in signals.py
#receiver(post_save, sender=_M)
def _p(sender, instance, created, **kwargs):
user = instance.user
No ?
Request object can be obtained from frame record by inspecting.
import inspect
request = [
frame_record[0].f_locals["request"]
for frame_record in inspect.stack()
if frame_record[3] == "get_response"
][0]
def get_requested_user():
import inspect
for frame_record in inspect.stack():
if frame_record[3] == 'get_response':
request = frame_record[0].f_locals['request']
return request.user
else:
return None
context_processors.py
from django.core.cache import cache
def global_variables(request):
cache.set('user', request.user)
----------------------------------
in you model
from django.db.models.signals import pre_delete
from django.dispatch import receiver
from django.core.cache import cache
from news.models import News
#receiver(pre_delete, sender=News)
def news_delete(sender, instance, **kwargs):
user = cache.get('user')
in settings.py
TEMPLATE_CONTEXT_PROCESSORS = (
'web.context_processors.global_variables',
)
I'm trying to make a custom form field in Django that allows a user to link a file or upload a file. To do this, I'm creating a subclass of the MultiValueField with the fields property set to (URLField(), FileField()). I'm not sure this is the right approach, but I keep getting an error I can't understand:
'MyFileField' object has no attribute 'attrs'
Here is my code. Can anyone explain what's going on?
from django import forms
from django.core.exceptions import ValidationError
from .models import Case, Person, Opp, Category
class MyFileField(forms.MultiValueField):
def compress(self, data_list):
# I'll get to this once the attr error goes away
pass
def __init__(self, *args, **kwargs):
fields = (
forms.URLField(), forms.FileField()
)
super(MyFileField, self).__init__(fields=fields, *args, **kwargs)
class CaseForm(forms.ModelForm):
class Meta:
model = Case
fields = ['title', 'file', 'categories']
widgets = {
'file': MyFileField
}
The problem is that you are calling init on an abstract class.
super(MyFileField, self).__init__(fields=fields, *args, **kwargs)
But the base class is abstract.
Look at https://docs.djangoproject.com/en/1.8/ref/forms/fields/#multivaluefield
and
Python's super(), abstract base classes, and NotImplementedError
I implemented a basic rest api with the django rest framework. It works perfectly using the browsable api or communicating to it with requests. Next step would be submitting data to the rest api.
Here is what I have done so far.
settings.py
REST_FRAMEWORK = {
'DEFAULT_PERMISSION_CLASSES': ('rest_framework.permissions.AllowAny',),
'PAGINATE_BY': 10
}
[UPDATE:]
models.py
class Request(models.Model):
name = models.TextField()
def save(self, *args, **kwargs):
super(Request, self).save(*args, **kwargs) # Call the "real" save() method.
serializers.py
class RequestSerializer(serializers.HyperlinkedModelSerializer):
class Meta:
model = Request
fields = ('id', 'name')
views.api
class RequestsViewSet(viewsets.ModelViewSet):
queryset = Request.objects.all()
serializer_class = RequestSerializer
Using the browsable api I see that those are the options supported:
Allow: GET, HEAD, OPTIONS
Obviously, POST (and also PUT) is missing.
What I am doing wrong?
Thanks!
Solved it by adding the post method to the modelviewset (in the view):
def post(self, request, format=None):
...
Thanks for helping!
Well, I think you only need to call save method on the model object to persist the object in the database.
First, import model to the view, instantiate a model object in the view, then call save method on the newly created object. If you have model connected to the backend, that will persist your changes.
models.py
class YourModel(models.Model):
name = models.CharField()
views.py
from models import YourModel
def yourView(request):
yourObject = YourModel(name='John')
yourObject.save()
...
Check also Django documentation for models here