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Extend django-import-export's import form to specify fixed value for each imported row

I am using django-import-export 1.0.1 with admin integration in Django 2.1.1. I have two models
from django.db import models
class Sector(models.Model):
code = models.CharField(max_length=30, primary_key=True)
class Location(models.Model):
code = models.CharField(max_length=30, primary_key=True)
sector = ForeignKey(Sector, on_delete=models.CASCADE, related_name='locations')
and they can be imported/exported just fine using model resources
from import_export import resources
from import_export.fields import Field
from import_export.widgets import ForeignKeyWidget
class SectorResource(resources.ModelResource):
code = Field(attribute='code', column_name='Sector')
class Meta:
model = Sector
import_id_fields = ('code',)
class LocationResource(resources.ModelResource):
code = Field(attribute='code', column_name='Location')
sector = Field(attribute='sector', column_name='Sector',
widget=ForeignKeyWidget(Sector, 'code'))
class Meta:
model = Location
import_id_fields = ('code',)
and import/export actions can be integrated into the admin by
from django.contrib import admin
from import_export.admin import ImportExportModelAdmin
class SectorAdmin(ImportExportModelAdmin):
resource_class = SectorResource
class LocationAdmin(ImportExportModelAdmin):
resource_class = LocationResource
admin.site.register(Sector, SectorAdmin)
admin.site.register(Location, LocationAdmin)
For Reasons™, I would like to change this set-up so that a spreadsheet of Locations which does not contain a Sector column can be imported; the value of sector (for each imported row) should be taken from an extra field on the ImportForm in the admin.
Such a field can indeed be added by overriding import_action on the ModelAdmin as described in Extending the admin import form for django import_export. The next step, to use this value for all imported rows, is missing there, and I have not been able to figure out how to do it.

EDIT(2): Solved through the use of sessions. Having a get_confirm_import_form hook would still really help here, but even better would be having the existing ConfirmImportForm carry across all the submitted fields & values from the initial import form.
EDIT: I'm sorry, I thought I had this nailed, but my own code wasn't working as well as I thought it was. This doesn't solve the problem of passing along the sector form field in the ConfirmImportForm, which is necessary for the import to complete. Currently looking for a solution which doesn't involve pasting the whole of import_action() into an ImportMixin subclass. Having a get_confirm_import_form() hook would help a lot here.
Still working on a solution for myself, and when I have one I'll update this too.
Don't override import_action. It's a big complicated method that you don't want to replicate. More importantly, as I discovered today: there are easier ways of doing this.
First (as you mentioned), make a custom import form for Location that allows the user to choose a Sector:
class LocationImportForm(ImportForm):
sector = forms.ModelChoiceField(required=True, queryset=Sector.objects.all())
In the Resource API, there's a before_import_row() hook that is called once per row. So, implement that in your LocationResource class, and use it to add the Sector column:
def before_import_row(self, row, **kwargs):
sector = self.request.POST.get('sector', None)
if contract:
self.request.session['import_context_sector'] = sector
else:
# if this raises a KeyError, we want to know about it.
# It means that we got to a point of importing data without
# contract context, and we don't want to continue.
try:
sector = self.request.session['import_context_sector']
except KeyError as e:
raise Exception("Sector context failure on row import, " +
f"check resources.py for more info: {e}")
row['sector'] = sector
(Note: This code uses Django sessions to carry the sector value from the import form to the import confirmation screen. If you're not using sessions, you'll need to find another way to do it.)
This is all you need to get the extra data in, and it works for both the dry-run preview and the actual import.
Note that self.request doesn't exist in the default ModelResource - we have to install it by giving LocationResource a custom constructor:
def __init__(self, request=None):
super()
self.request = request
(Don't worry about self.request sticking around. Each LocationResource instance doesn't persist beyond a single request.)
The request isn't usually passed to the ModelResource constructor, so we need to add it to the kwargs dict for that call. Fortunately, Django Import/Export has a dedicated hook for that. Override ImportExportModelAdmin's get_resource_kwargs method in LocationAdmin:
def get_resource_kwargs(self, request, *args, **kwargs):
rk = super().get_resource_kwargs(request, *args, **kwargs)
rk['request'] = request
return rk
And that's all you need.

how to access field using django signals

I would like to check the value of a field (lightStatusA) when a new record is saved into my Django database. I feel like iv'e read the docs 10 times and still can't grasp how to get this. Here is my current models.py code:
from django.db import models
from accounts.models import Customer
from django.conf import settings
from django.contrib.auth import get_user_model
from django.db.models.signals import post_save
class Data(models.Model):
author = models.ForeignKey(get_user_model(),on_delete=models.CASCADE,)
tempData= models.CharField(max_length=50,blank=True,)
humidData= models.CharField(max_length=50,blank=True,)
lightStatusA= models.BooleanField(default=True,)
dateTime = models.DateTimeField(auto_now_add=True)
def __str__(self):
return str(self.author)
def check_light_status(sender, **kwargs):
if kwargs['created']: #make sure its a new record
#need logic to grab instance.lightStatusA and check it's value
post_save.connect(check_light_status, sender = Data)
Is there some way of passing this value as an argument to the signal?

The check_light_status function can accept an instance parameter.
From the docs:
instance
The actual instance being saved.
Update: You said this:
instance returns the author of the post.
I am going to use my powers of deduction to guess that you tried print(instance) and saw the author. Look at your __str__ implementation.
def __str__(self):
return str(self.author)
I'd say you sabotaged yourself a bit there ;)

Ahh ok got it:
def check_light_status(sender, instance, **kwargs):
if kwargs['created']: #make sure its a new record
print(instance.lightStatusA)
This prints the the field I need to run some logic against.

Single model dynamic database settings in Django

For example assume that I have 100 clients who uses WordPress and I have to write a service in Django which should return list of posts from WordPress's MySQL DB. The problem is 100 clients are having different database connection settings.
I know that I can use DatabaseRouter to switch databases which are already loaded in settings. But I don't know how to make a singe model class to use different database settings.
I have tried mutating settings.
I also tried mutating model's app_label.
But I later understood that mutating anyting in Django is meaning less.
My Requirements
I want to create a model and dynamically change database connection. List of connection can be in a managed database table. But I don't want to unnecessarily load all the connection settings or create multiple models.

I made something like that, but to change mongodb connections.
I created a GenericView that select the connection and use it on the get_queryset.
I'm using django rest framework, so I made something like this:
class SwitchDBMixinView(object):
model = None
fields = None
def initial(self, request, *args, **kwargs):
result = super().initial(request, *args, **kwargs)
if request.user.is_authenticated():
request.user.database_connection.register()
return result
def get_object(self, *args, **kwargs):
return super().get_object(*args, **kwargs).switch_db(self.get_db_alias())
def get_db_alias(self):
if self.request is None or not self.request.user.is_authenticated():
return DEFAULT_CONNECTION_NAME
return self.request.user.database_connection.name
def get_queryset(self):
return self.model.objects.using(self.get_db_alias()).all()
def perform_destroy(self, instance):
instance.switch_db(self.get_db_alias()).delete()
The model:
from mongoengine.connection import register_connection, get_connection
AUTH_USER_MODEL = getattr(settings, 'AUTH_USER_MODEL')
class Connection(models.Model):
class Meta:
pass
owner = models.OneToOneField(
AUTH_USER_MODEL,
related_name='database_connection',
)
uri = models.TextField(
default=DefaultMongoURI()
)
def register(self):
register_connection(
self.name,
host=self.uri,
tz_aware=True,
)
get_connection(
self.name,
reconnect=True
)
def get_name(self):
return 'client-%d' % self.owner.pk
name = property(get_name)
def __str__(self):
return self.uri

You may want to have a look at django.db.connections (in django/db/__init__.py) and django.db.utils.ConnectionHandler (which django.db.connections is an instance of). This should let you dynamically add new db configs without hacking settings.DATABASES (actually ConnectionHandler builds it's _databases attribute from settings.DATABASES). I can't tell for sure since I never tried but it should mostly boils down to
from django import db
def add_db(alias, connection_infos):
databases = db.connections.databases
if alias in databases:
either_raise_or_log_and_ignore(your choice)
db.connections.databases[alias] = connection_infos
where connection_infos is a mapping similar to the ones expected in settings.DATABASES.
Then it's mostly a matter of using Queryset.using(alias) for your queries, ie:
alias = get_alias_for_user(request.user)
posts = Post.objects.using(alias).all()
cf https://docs.djangoproject.com/en/1.11/topics/db/multi-db/#manually-selecting-a-database
The main problem with this IMHO (assuming you manage to make something that works out of the untested suggestion above) is that you will have to store databases users/password in clear somewhere which can be a major security issue. I don't know how much control you have on the databases admin part but it would be better if you could add a 'django' user with a same password (and appropriate permissions of course) on all those databases so you can keep the password in your settings file instead of having to keep it in your main db.

Accessing the user's request in a post_save signal

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',
)

Django models: Only permit one entry in a model?

I want to make some of my Django global settings configurable through the admin interface.
To that end, I've decided to set them as database fields, rather than in settings.py.
These are the settings I care about:
class ManagementEmail(models.Model):
librarian_email = models.EmailField()
intro_text = models.CharField(max_length=1000)
signoff_text = models.CharField(max_length=1000)
These are one-off global settings, so I only ever want there to be a single librarian_email, intro_text etc floating around the system.
Is there a way I can prevent admin users from adding new records here, without preventing them from editing the existing record?
I guess I can do this by writing a custom admin template for this model, but I'd like to know if there's a neater way to configure this.
Could I use something other than class, for example?
Thanks!

Please see this question on "keep[ing] settings in database", where the answer seems to be django-dbsettings
Update
Just thought of another option: you can create the following model:
from django.contrib.sites.models import Site
class ManagementEmail(models.Model):
site = models.OneToOneField(Site)
librarian_email = models.EmailField()
intro_text = models.CharField(max_length=1000)
signoff_text = models.CharField(max_length=1000)
Because of the OneToOneField field, you can only have one ManagementEmail record per site. Then, just make sure you're using sites and then you can pull the settings thusly:
from django.contrib.sites.models import Site
managementemail = Site.objects.get_current().managementemail
Note that what everyone else is telling you is true; if your goal is to store settings, adding them one by one as fields to a model is not the best implementation. Adding settings over time is going to be a headache: you have to add the field to your model, update the database structure, and modify the code that is calling that setting.
That's why I'd recommend using the django app I mentioned above, since it does exactly what you want -- provide for user-editable settings -- without making you do any extra, unnecessary work.

I think the easiest way you can do this is using has_add_permissions function of the ModelAdmin:
class ContactUsAdmin(admin.ModelAdmin):
form = ContactUsForm
def has_add_permission(self, request):
return False if self.model.objects.count() > 0 else super().has_add_permission(request)
You can set the above to be any number you like, see the django docs.
If you need more granularity than that, and make the class a singleton at the model level, see django-solo. There are many singleton implementations also that I came across.
For StackedInline, you can use max_num = 1.

Try django-constance.
Here are some useful links:
https://github.com/jezdez/django-constance
http://django-constance.readthedocs.org/en/latest/

I'd take a page out of wordpress and create a Model that support settings.
class Settings(models.Model):
option_name = models.CharField(max_length=1000)
option_value = models.CharField(max_length=25000)
option_meta = models.CharField(max_length=1000)
Then you can just pickle (serialize) objects into the fields and you'll be solid.
Build a little api, and you can be as crafty as wordpress and call. AdminOptions.get_option(opt_name)
Then you can just load the custom settings into the runtime, keeping the settings.py module separate, but equal. A good place to write this would be in an __init__.py file.

Just set up an GlobalSettings app or something with a Key and Value field.
You could easily prevent admin users from changing values by not giving them permission to edit the GlobalSettings app.
class GlobalSettingsManager(models.Manager):
def get_setting(self, key):
try:
setting = GlobalSettings.objects.get(key=key)
except:
raise MyExceptionOrWhatever
return setting
class GlobalSettings(models.Model):
key = models.CharField(unique=True, max_length=255)
value = models.CharField(max_length=255)
objects = GlobalSettingsManager()
>>> APP_SETTING = GlobalSettings.objects.get_setting('APP_SETTING')
There are apps for this but I prefer looking at them and writing my own.

You can prevent users from adding/deleting an object by overriding this method on your admin class:
ModelAdmin.has_add_permission(self, request)
ModelAdmin.has_delete_permission(self, request, obj=None)

Modification of #radtek answer to prevent deleting if only one entry is left
class SendgridEmailQuotaAdmin(admin.ModelAdmin):
list_display = ('quota','used')
def has_add_permission(self, request):
return False if self.model.objects.count() > 0 else True
def has_delete_permission(self, request, obj=None):
return False if self.model.objects.count() <= 1 else True
def get_actions(self, request):
actions = super(SendgridEmailQuotaAdmin, self).get_actions(request)
if(self.model.objects.count() <= 1):
del actions['delete_selected']
return actions

I had basically the same problem as the original poster describes, and it's easily fixed by overriding modelAdmin classes. Something similar to this in an admin.py file easily prevents adding a new object but allows editing the current one:
class TitleAdmin(admin.ModelAdmin):
def has_delete_permission(self, request, obj=TestModel.title):
return False
def has_add_permission(self, request):
return False
def has_change_permission(self, request, obj=TestModel.title):
return True
This doesn't prevent a user from posting a form that edits data, but keeps things from happening in the Admin site. Depending on whether or not you feel it's necessary for your needs you can enable deletion and addition of a record with a minimum of coding.