I have a model that saves an Excursion. The user can change this excursion, but I need to know what the excursion was before he change it, because I keep track of how many "bookings" are made per excursion, and if you change your excursion, I need to remove one booking from the previous excursion.
Im not entirely sure how this should be done.
Im guessing you use a signal for this?
Should I use pre_save, pre_init or what would be the best for this?
pre_save is not the correct one it seems, as it prints the new values, not the "old value" as I expected
#receiver(pre_save, sender=Delegate)
def my_callback(sender, instance, *args, **kwargs):
print instance.excursion
Do you have several options.
First one is to overwrite save method:
#Delegate
def save(self, *args, **kwargs):
if self.pk:
previous_excursion = Delegate.objects.get(self.pk).excursion
super(Model, self).save(*args, **kwargs)
if self.pk and self.excursion != previous_excursion:
#change booking
Second one is binding function to post save signal + django model utils field tracker:
#receiver(post_save, sender=Delegate)
def create_change_booking(sender,instance, signal, created, **kwargs):
if created:
previous_excursion = get it from django model utils field tracker
#change booking
And another solution is in pre_save as you are running:
#receiver(pre_save, sender=Delegate)
def my_callback(sender, instance, *args, **kwargs):
previous_excursion = Delegate.objects.get(self.pk).excursion
if instance.pk and instance.excursion != previous_excursion:
#change booking
You can use django model utils to track django model fields. check this example.
pip install django-model-utils
Then you can define your model and use fieldtracker in your model .
from django.db import models
from model_utils import FieldTracker
class Post(models.Model):
title = models.CharField(max_length=100)
body = models.TextField()
tracker = FieldTracker()
status = models.CharField(choices=STATUS, default=STATUS.draft, max_length=20)
after that in post save you can use like this :
#receiver(post_save, sender=Post)
def my_callback(sender, instance,*args, **kwargs):
print (instance.title)
print (instance.tracker.previous('title'))
print (instance.status)
print (instance.tracker.previous('status'))
This will help you a lot to do activity on status change. as because overwrite save method is not good idea.
As an alternative and if you are using Django forms:
The to-be version of your instance is stored in form.instance of the Django form of your model. On save, validations are run and this new version is applied to the model and then the model is saved.
Meaning that you can check differences between the new and the old version by comparing form.instance to the current model.
This is what happens when the Django Admin's save_model method is called. (See contrib/admin/options.py)
If you can make use of Django forms, this is the most Djangothic way to go, I'd say.
This is the essence on using the Django form for handling data changes:
form = ModelForm(request.POST, request.FILES, instance=obj)
new_object = form.instance # not saved yet
# changes are stored in form.changed_data
new_saved_object = form.save()
form.changed_data will contain the changed fields which means that it is empty if there are no changes.
There's yet another option:
Django's documentation has an example showing exactly how you could do this by overriding model methods.
In short:
override Model.from_db() to add a dynamic attribute containing the original values
override the Model.save() method to compare the new values against the originals
This has the advantage that it does not require an additional database query.
Related
I would like to control some configuration settings for my project using a database model. For example:
class JuicerBaseSettings(models.Model):
max_rpm = model.IntegerField(default=10)
min_rpm = model.IntegerField(default=0)
There should only be one instance of this model:
juicer_base = JuicerBaseSettings()
juicer_base.save()
Of course, if someone accidentally creates a new instances, it's not the end of the world. I could just do JuicerBaseSettings.objects.all().first(). However, is there a way to lock it down such that it's impossible to create more than 1 instance?
I found two related questions on SO. This answer suggests using 3rd party apps like django-singletons, which doesn't seem to be actively maintained (last update to the git repo is 5 years ago). Another answer suggests using a combination of either permissions or OneToOneField. Both answers are from 2010-2011.
Given that Django has changed a lot since then, are there any standard ways to solve this problem? Or should I just use .first() and accept that there may be duplicates?
You can override save method to control number of instances:
class JuicerBaseSettings(models.Model):
def save(self, *args, **kwargs):
if not self.pk and JuicerBaseSettings.objects.exists():
# if you'll not check for self.pk
# then error will also raised in update of exists model
raise ValidationError('There is can be only one JuicerBaseSettings instance')
return super(JuicerBaseSettings, self).save(*args, **kwargs)
Either you can override save and create a class function JuicerBaseSettings.object()
class JuicerBaseSettings(models.Model):
#classmethod
def object(cls):
return cls._default_manager.all().first() # Since only one item
def save(self, *args, **kwargs):
self.pk = self.id = 1
return super().save(*args, **kwargs)
============= OR =============
Simply, Use django_solo.
https://github.com/lazybird/django-solo
Snippet Courtsy: django-solo-documentation.
# models.py
from django.db import models
from solo.models import SingletonModel
class SiteConfiguration(SingletonModel):
site_name = models.CharField(max_length=255, default='Site Name')
maintenance_mode = models.BooleanField(default=False)
def __unicode__(self):
return u"Site Configuration"
class Meta:
verbose_name = "Site Configuration"
# admin.py
from django.contrib import admin
from solo.admin import SingletonModelAdmin
from config.models import SiteConfiguration
admin.site.register(SiteConfiguration, SingletonModelAdmin)
# There is only one item in the table, you can get it this way:
from .models import SiteConfiguration
config = SiteConfiguration.objects.get()
# get_solo will create the item if it does not already exist
config = SiteConfiguration.get_solo()
If your model is used in django-admin only, you additionally can set dynamic add permission for your model:
# some imports here
from django.contrib import admin
from myapp import models
#admin.register(models.ExampleModel)
class ExampleModelAdmin(admin.ModelAdmin):
# some code...
def has_add_permission(self, request):
# check if generally has add permission
retVal = super().has_add_permission(request)
# set add permission to False, if object already exists
if retVal and models.ExampleModel.objects.exists():
retVal = False
return retVal
i am not an expert but i guess you can overwrite the model's save() method so that it will check if there has already been a instance , if so the save() method will just return , otherwise it will call the super().save()
You could use a pre_save signal
#receiver(pre_save, sender=JuicerBaseSettings)
def check_no_conflicting_juicer(sender, instance, *args, **kwargs):
# If another JuicerBaseSettings object exists a ValidationError will be raised
if JuicerBaseSettings.objects.exclude(pk=instance.pk).exists():
raise ValidationError('A JuiceBaseSettings object already exists')
I'm a bit late to the party but if you want to ensure that only one instance of an object is created, an alternative solution to modifying a models save() function would be to always specify an ID of 1 when creating an instance - that way, if an instance already exists, an integrity error will be raised.
e.g.
JuicerBaseSettings.objects.create(id=1)
instead of:
JuicerBaseSettings.objects.create()
It's not as clean of a solution as modifying the save function but it still does the trick.
I did something like this in my admin so that I won't ever go to original add_new view at all unless there's no object already present:
def add_view(self, request, form_url='', extra_context=None):
obj = MyModel.objects.all().first()
if obj:
return self.change_view(request, object_id=str(obj.id) if obj else None)
else:
return super(type(self), self).add_view(request, form_url, extra_context)
def changelist_view(self, request, extra_context=None):
return self.add_view(request)
Works only when saving from admin
I have a signal inside my django app where I would like to check if a certain field in my model has been updated, so I can then proceed and do something.
My model looks like this...
class Product(models.Model):
name = models.CharField(max_length=100)
price = models.PositiveIntegerField()
tax_rate = models.PositiveIntegerField()
display_price = models.PositiveInteger()
inputed_by = models.ForeignKey(settings.AUTH_USER_MODEL, null=True, blank=True, on_delete=models.SET_NULL)
updated_by = models.ForeignKey(settings.AUTH_USER_MODEL, null=True, blank=True, on_delete=models.SET_NULL)
My signal looks like this...
#receiver(post_save, sender=Product)
def update_model(sender, **kwargs):
instance = kwargs['instance']
if 'tax_rate' in kwargs['update_fields']:
# do something
This returns the error None is not an iterable. I have read the django signal documentation regarding the update_fields and it says The set of fields to update as passed to Model.save(), or None if update_fields wasn’t passed to save().
I should mention that I am working inside django admin here so what I hoped would happen is, I could create an instance of my Product model in django admin and then later if the value of tax_rate or price were updated, I could check for those and update the list_price accordingly. However, kwargs['update_fields'] always returns None.
What am I getting wrong? Or is there some other way I could achieve that result inside django admin?
Updated section
Now, say I introduce a field called inputed_by in my product model, that points to the user model and I want that field populated when the model is first saved. Then another field updated_by that stores the user who last updated the model. At the same time I wish to check whether either or both the tax_rate or price has been updated.
Inside my model admin I have the following method...
def save_model(self, request, obj, form, change):
update_fields = []
if not obj.pk:
obj.inputed_by = request.user
elif change:
obj.updated_by = request.user
if form.initial['tax_rate'] != form.cleaned_data['tax_rate']:
update_fields.append('tax_rate')
if form.initial['price'] != form.cleaned_data['price']:
update_fields.append('price')
obj.save(update_fields=update_fields)
super().save_model(request, obj, form, change)
My signal now looks like this...
#receiver(post_save, sender=Product, dispatch_uid="update_display_price")
def update_display_price(sender, **kwargs):
created = kwargs['created']
instance = kwargs['instance']
updated = kwargs['update_fields']
checklist = ['tax_rate', 'price']
# Prints out the frozenset containing the updated fields and then below that `The update_fields is None`
print(f'The update_fields is {updated}')
if created:
instance.display_price = instance.price+instance.tax_rate
instance.save()
elif set(checklist).issubset(updated):
instance.display_price = instance.price+instance.tax_rate
instance.save()
I get the error 'NoneType' object is not iterable
The error seems to come from the line set(checklist).issubset(updated). I've tried running that line specifically inside the python shell and it yields the desired results. What's wrong this time?
The set of fields should be passed to Model.save() to make them available in update_fields.
Like this
model.save(update_fields=['tax_rate'])
If you are creating something from django admin and getting always None it means that update_fields has not been passed to model's save method. And because of that it will always be None.
If you check ModelAdmin class and save_model method you'll see that call happens without update_fields keyword argument.
It will work if you write your own save_model.
The code below will solve your problem:
class ProductAdmin(admin.ModelAdmin):
...
def save_model(self, request, obj, form, change):
update_fields = []
# True if something changed in model
# Note that change is False at the very first time
if change:
if form.initial['tax_rate'] != form.cleaned_data['tax_rate']:
update_fields.append('tax_rate')
obj.save(update_fields=update_fields)
Now you'll be able to test memberships in update_model.
To add to Davit Tovmasyan's post. I made a more universal version that covers any field change using a for loop:
class ProductAdmin(admin.ModelAdmin):
...
def save_model(self, request, obj, form, change):
update_fields = []
for key, value in form.cleaned_data.items():
# True if something changed in model
if value != form.initial[key]:
update_fields.append(key)
obj.save(update_fields=update_fields)
EDIT: WARNING This isnt actually a full solution. Doesnt seem to work for object creation, only changes. I will try to figure out the full solution soon.
I wanted to add an alternative that relies on the pre_save signal to get the previous version of the instance you're evaluating (from this SO answer):
#receiver(pre_save, sender=Product)
def pre_update_model(sender, **kwargs):
# check if the updated fields exist and if you're not creating a new object
if not kwargs['update_fields'] and kwargs['instance'].id:
# Save it so it can be used in post_save
kwargs['instance'].old = User.objects.get(id=kwargs['instance'].id)
#receiver(post_save, sender=Product)
def update_model(sender, **kwargs):
instance = kwargs['instance']
# Add updated_fields, from old instance, so the method logic remains unchanged
if not kwargs['update_fields'] and hasattr(instance, 'old'):
kwargs['update_fields'] = []
if (kwargs['update_fields'].instance.tax_rate !=
kwargs['update_fields'].instance.old.tax_rate):
kwargs['update_fields'].append('tax_rate')
if 'tax_rate' in kwargs['update_fields']:
comparing to the accepted answer
Disadvantages
Extra query on every save that doesn't have update_fields (if you're not opening Django Admin to the world, this shouldn't be problematic)
Advantages
Don't need to override any method or class
You only need to implement the logic for the fields you want to evaluate, and they are in the same method, so no excuse for mistakes ;)
If you're doing this for many classes, you should probably look at other solutions (but the accepted answer is also not perfect for that!)
You can do this.
def save_model(self, request, obj, form, change):
if change:
obj.save(update_fields=form.changed_data)
else:
super().save_model(request, obj, form, change)
If it is possible to do a post save based on a field change, would the code look something like this? Note: I want a Car object to get created each time the address of the House model changes.
def create_car(sender, **kwargs):
if kwargs['??????']:
car = Car.objects.filter(user=kwargs['instance'].user)
post_save.connect(create_car, sender=House.address)
I'm not sure what would go in the kwargs.
Thanks!
You can compare address fields in pre_save signal
#receiver(signals.pre_save, sender=House)
def create_car(sender, instance, **kwargs):
try:
old_instance = sender.objects.get(pk=instance.pk)
except sender.DoesNotExist:
return
if instance.address != old_instance.address:
car = Car.objects.filter(user=kwargs['instance'].user)
# todo
I need to pass an instance variable (self.rank) to be used by a class variable (provider) (see the commented out line below).
Commented out, the code below works. But I'm pretty sure I shouldn't be trying to pass an instance variable up to a class variable anyway. So I'm dumbfounded as to how to accomplish my goal, which is to dynamically filter down my data in the ModelChoiceField.
As you can see, I already overrided ModelChoiceField so I could beautify the usernames. And I also subclassed my basic SwapForm because I have several other forms I'm using (not shown here).
Another way of saying what I need ... I want the value of request.user in my Form so I can then determine the rank of that user and then filter out my Users by rank to build a smaller ModelChoiceField (that looks good too). Note that in my views.py, I call the form using:
form = NewSwapForm(request.user)
or
form = NewSwapForm(request.user, request.POST)
In forms.py:
from myapp.swaps.models import Swaps
from django.contrib.auth.models import User
class UserModelChoiceField(forms.ModelChoiceField):
""" Override the ModelChoiceField to display friendlier name """
def label_from_instance(self, obj):
return "%s" % (obj.get_full_name())
class SwapForm(forms.ModelForm):
""" Basic form from Swaps model. See inherited models below. """
class Meta:
model = Swaps
class NewSwapForm(SwapForm):
# Using a custom argument 'user'
def __init__(self, user, *args, **kwargs):
super(NewSwapForm, self).__init__(*args, **kwargs)
self.rank = User.objects.get(id=user.id).firefighter_rank_set.get().rank
provider = UserModelChoiceField(User.objects.all().
order_by('last_name').
filter(firefighter__hirestatus='active')
### .filter(firefighter_rank__rank=self.rank) ###
)
class Meta(SwapForm.Meta):
model = Swaps
fields = ['provider', 'date_swapped', 'swap_shift']
Thanks!
You can't do it that way, because self doesn't exist at that point - and even if you could, that would be executed at define time, so the rank would be static for all instantiations of the form.
Instead, do it in __init__:
provider = UserModelChoiceField(User.objects.none())
def __init__(self, user, *args, **kwargs):
super(NewSwapForm, self).__init__(*args, **kwargs)
rank = User.objects.get(id=user.id).firefighter_rank_set.get().rank # ??
self.fields['provider'].queryset = User.objects.order_by('last_name').filter(
firefighter__hirestatus='active', firefighter_rank__rank=rank)
I've put a question mark next to the rank line, because rank_set.get() isn't valid... not sure what you meant there.
I suppose similar problem would have been discussed here, but I couldn't find it.
Let's suppose I have an Editor and a Supervisor. I want the Editor to be able to add new content (eg. a news post) but before publication it has to be acknowledged by Supervisor.
When Editor lists all items, I want to set some fields on the models (like an 'ack' field) as read-only (so he could know what had been ack'ed and what's still waiting approval) but the Supervisor should be able to change everything (list_editable would be perfect)
What are the possible solutions to this problem?
I think there is a more easy way to do that:
Guest we have the same problem of Blog-Post
blog/models.py:
Class Blog(models.Model):
...
#fields like autor, title, stuff..
...
class Post(models.Model):
...
#fields like blog, title, stuff..
...
approved = models.BooleanField(default=False)
approved_by = models.ForeignKey(User)
class Meta:
permissions = (
("can_approve_post", "Can approve post"),
)
And the magic is in the admin:
blog/admin.py:
...
from django.views.decorators.csrf import csrf_protect
...
def has_approval_permission(request, obj=None):
if request.user.has_perm('blog.can_approve_post'):
return True
return False
Class PostAdmin(admin.ModelAdmin):
#csrf_protect
def changelist_view(self, request, extra_context=None):
if not has_approval_permission(request):
self.list_display = [...] # list of fields to show if user can't approve the post
self.editable = [...]
else:
self.list_display = [...] # list of fields to show if user can approve the post
return super(PostAdmin, self).changelist_view(request, extra_context)
def get_form(self, request, obj=None, **kwargs):
if not has_approval_permission(request, obj):
self.fields = [...] # same thing
else:
self.fields = ['approved']
return super(PostAdmin, self).get_form(request, obj, **kwargs)
In this way you can use the api of custom permission in django, and you can override the methods for save the model or get the queryset if you have to. In the methid has_approval_permission you can define the logic of when the user can or can't to do something.
Starting Django 1.7, you can now use the get_fields hook which makes it so much simpler to implement conditional fields.
class MyModelAdmin(admin.ModelAdmin):
...
def get_fields(self, request, obj=None):
fields = super(MyModelAdmin, self).get_fields(request, obj)
if request.user.is_superuser:
fields += ('approve',)
return fields
I have a system kind of like this on a project that I'm just finishing up. There will be a lot of work to put this together, but here are some of the components that I had to make my system work:
You need a way to define an Editor and a Supervisor. The three ways this could be done are 1.) by having an M2M field that defines the Supervisor [and assuming that everyone else with permission to read/write is an Editor], 2.) make 2 new User models that inherit from User [probably more work than necessary] or 3.) use the django.auth ability to have a UserProfile class. Method #1 is probably the most reasonable.
Once you can identify what type the user is, you need a way to generically enforce the authorization you're looking for. I think the best route here is probably a generic admin model.
Lastly you'll need some type of "parent" model that will hold the permissions for whatever needs to be moderated. For example, if you had a Blog model and BlogPost model (assuming multiple blogs within the same site), then Blog is the parent model (it can hold the permissions of who approves what). However, if you have a single blog and there is no parent model for BlogPost, we'll need some place to store the permissions. I've found the ContentType works out well here.
Here's some ideas in code (untested and more conceptual than actual).
Make a new app called 'moderated' which will hold our generic stuff.
moderated.models.py
class ModeratedModelParent(models.Model):
"""Class to govern rules for a given model"""
content_type = models.OneToOneField(ContentType)
can_approve = models.ManyToManyField(User)
class ModeratedModel(models.Model):
"""Class to implement a model that is moderated by a supervisor"""
is_approved = models.BooleanField(default=False)
def get_parent_instance(self):
"""
If the model already has a parent, override to return the parent's type
For example, for a BlogPost model it could return self.parent_blog
"""
# Get self's ContentType then return ModeratedModelParent for that type
self_content_type = ContentType.objects.get_for_model(self)
try:
return ModeratedModelParent.objects.get(content_type=self_content_type)
except:
# Create it if it doesn't already exist...
return ModeratedModelParent.objects.create(content_type=self_content_type).save()
class Meta:
abstract = True
So now we should have a generic, re-usable bit of code that we can identify the permission for a given model (which we'll identify the model by it's Content Type).
Next, we can implement our policies in the admin, again through a generic model:
moderated.admin.py
class ModeratedModelAdmin(admin.ModelAdmin):
# Save our request object for later
def __call__(self, request, url):
self.request = request
return super(ModeratedModelAdmin, self).__call__(request, url)
# Adjust our 'is_approved' widget based on the parent permissions
def formfield_for_dbfield(self, db_field, **kwargs):
if db_field.name == 'is_approved':
if not self.request.user in self.get_parent_instance().can_approve.all():
kwargs['widget'] = forms.CheckboxInput(attrs={ 'disabled':'disabled' })
# Enforce our "unapproved" policy on saves
def save_model(self, *args, **kwargs):
if not self.request.user in self.get_parent_instance().can_approve.all():
self.is_approved = False
return super(ModeratedModelAdmin, self).save_model(*args, **kwargs)
Once these are setup and working, we can re-use them across many models as I've found once you add structured permissions for something like this, you easily want it for many other things.
Say for instance you have a news model, you would simply need to make it inherit off of the model we just made and you're good.
# in your app's models.py
class NewsItem(ModeratedModel):
title = models.CharField(max_length=200)
text = models.TextField()
# in your app's admin.py
class NewsItemAdmin(ModeratedModelAdmin):
pass
admin.site.register(NewsItem, NewsItemAdmin)
I'm sure I made some code errors and mistakes in there, but hopefully this can give you some ideas to act as a launching pad for whatever you decide to implement.
The last thing you have to do, which I'll leave up to you, is to implement filtering for the is_approved items. (ie. you don't want un-approved items being listed on the news section, right?)
The problem using the approach outlined by #diegueus9 is that the ModelAdmin acts liked a singleton and is not instanced for each request. This means that each request is modifying the same ModelAdmin object that is being accessed by other requests, which isn't ideal. Below is the proposed solutions by #diegueus9:
# For example, get_form() modifies the single PostAdmin's fields on each request
...
class PostAdmin(ModelAdmin):
def get_form(self, request, obj=None, **kwargs):
if not has_approval_permission(request, obj):
self.fields = [...] # list of fields to show if user can't approve the post
else:
self.fields = ['approved', ...] # add 'approved' to the list of fields if the user can approve the post
...
An alternative approach would be to pass fields as a keyword arg to the parent's get_form() method like so:
...
from django.contrib.admin.util import flatten_fieldsets
class PostAdmin(ModelAdmin):
def get_form(self, request, obj=None, **kwargs):
if has_approval_permission(request, obj):
fields = ['approved']
if self.declared_fieldsets:
fields += flatten_fieldsets(self.declared_fieldsets)
# Update the keyword args as needed to allow the parent to build
# and return the ModelForm instance you require for the user given their perms
kwargs.update({'fields': fields})
return super(PostAdmin, self).get_form(request, obj=None, **kwargs)
...
This way, you are not modifying the PostAdmin singleton on every request; you are simply passing the appropriate keyword args needed to build and return the ModelForm from the parent.
It is probably worth looking at the get_form() method on the base ModelAdmin for more info: https://code.djangoproject.com/browser/django/trunk/django/contrib/admin/options.py#L431