Develop Reference

How to have 2 Types with the same model in Graphene?

I created a model Checkout on my project, with a CheckoutType to handle the requests, but now i need a Profile, that is basically just getting many of the fields on Checkout. The problem is that Checkout and Profile will be retrieved by users with very different permissions, and the while the first one will have the right ones, the second one must not have them. so i went with creating 2 types:
Checkout:
class CheckoutType(ModelType):
class Meta:
model = Checkout
interfaces = [graphene.relay.Node]
connection_class = CountableConnection
permissions = ['app.view_checkout']
filter_fields = {
'zone': ['exact'],
'vehicle__mark': ['exact'],
'status': ['exact']
}
Profile:
class ProfileFilter(django_filters.FilterSet):
class Meta:
model = Checkout
fields = ['zone','status']
#property
def qs(self):
# The query context can be found in self.request.
return super(ProfileFilter, self).qs.filter(salesman=self.request.user)
class ProfileType(ModelType):
class Meta:
model = Checkout
interfaces = [graphene.relay.Node]
connection_class = CountableConnection
filterset_class = ProfileFilter
The thing here is that, the first one shouldn't filter, and just be a regular schema, while the second one should filter by the user that made the request, that and the permissions is the reason i use 2, but as soon as i implemented, all the tests i did for the Checkout Type started to fail, since it seems it tries to use the ProfileType. I searched a little, and it seems that relay only allows a type per model in Django, so this approach doesn't seems possible, but i'm not sure how to overwrite the CheckoutType on another schema, or how to make a second Type with different permissions and different filters. Does someone knows if this is possible?

Just in case someone is on the same boat, i think i found a way to make it work, but with a different approach, i just modified the CheckoutType a little:
class CheckoutType(ModelType):
# Meta
#classmethod
def get_queryset(cls, queryset, info):
if info.context.user.has_perm('app.view_checkout'):
return queryset
return queryset.filter(salesman=info.context.user)
Basically here i remove the permission from the Meta, since i don't want to check that there, and then i overwrite the get_queryset() to check if the user has the perms, if that's the case, then just return the normal query, but if not just filter(And any additional thing you want to do for people without the permission). I'm not sure if there's a better way, but definitely did the job.

Custom permissions on Django group form

I've added some custom permissions to my Post model.
I've also created a form to add/edit groups with only this custom permissions:
class GroupFornm(forms.ModelForm):
permissions = forms.MultipleChoiceField(choices=Post._meta.permissions)
class Meta:
model = Group
fields = '__all__'
It works because I can see and select only my custom permissions but when I try to save the form I got the error:
invalid literal for int() with base 10: 'can_view'
What am I doing wrong? It seems that this form field waits for (int, str) pair but documentation says that as usually, (str, str) should work.
Edit
Post._meta.permissions:
(('can_view', 'Can see tickets'), ('can_edit', 'Can edit tickets'), ('can_delete', 'Can delete tickets'), ('can_create', 'Can add new tickets'))

The problem is not really related to the form itself, but the fact that you somehow need to translate those permissions into Permission objects that should be stored in the Group instance (the one that this ModelForm is managing).
I think displaying the options is not a problem. But if a user later for example performs a POST request, with the options (like can_write), then the question is how the Form should translate these into Permission objects (or the primary keys of Permission objects).
In that case you need to coerce the name of the permissions to Permission objects, or the ids of Permission objects. We can for example use a TypedMultipleChoiceField, and coerce with:
def get_permission_from_name(name):
return Permission.objects.get(name=name)
class GroupFornm(forms.ModelForm):
permissions = forms.TypedMultipleChoiceField(
choices=Post._meta.permissions,
coerce=get_permission_from_name,
)
class Meta:
model = Group
fields = '__all__'
Note that the above is not really a very efficient implementation, since it requires a query for every value send. Furthermore in case no permission with that name exists, then this will raise an error.
If you want to construct Permissions on the fly (in case these are not yet constructed), then you can change the function to:
def get_permission_from_name(name):
return Permission.objects.get_or_create(
name=name,
defaults={
'content_type': ContentType.objects.get_for_model(Post),
'codename': name
}
)

Django Rest Framework updating nested m2m objects. Does anyone know a better way?

I have a case when user needs to update one instance together with adding/editing the m2m related objects on this instance.
Here is my solution:
# models.py
class AdditionalAction(SoftDeletionModel):
ADDITIONAL_CHOICES = (
('to_bring', 'To bring'),
('to_prepare', 'To prepare'),
)
title = models.CharField(max_length=50)
type = models.CharField(choices=ADDITIONAL_CHOICES, max_length=30)
class Event(models.Model):
title= models.CharField(max_length=255)
actions = models.ManyToManyField(AdditionalAction, blank=True)
# serializers.py
class MySerializer(serializers.ModelSerializer):
def update(self, instance, validated_data):
actions_data = validated_data.pop('actions')
# Use atomic block to rollback if anything raised Exception
with transaction.atomic():
# update main object
updated_instance = super().update(instance, validated_data)
actions = []
# Loop over m2m relation data and
# create/update each action instance based on id present
for action_data in actions_data:
action_kwargs = {
'data': action_data
}
id = action_data.get('id', False)
if id:
action_kwargs['instance'] = AdditionalAction.objects.get(id=id)
actions_ser = ActionSerializerWrite(**action_kwargs)
actions_ser.is_valid(raise_exception=True)
actions.append(actions_ser.save())
updated_instance.actions.set(actions)
return updated_instance
Can anyone suggest better solution?
P.S. actions can be created or updated in this case, so i can't just use many=True on serializer cause it also needs instance to update.

Using for loop with save here will be a killer if you have a long list or actions triggered on save, etc. I'd try to avoid it.
You may be better off using ORMS update with where clause: https://docs.djangoproject.com/en/2.0/topics/db/queries/#updating-multiple-objects-at-once and even reading the updated objects from the database after the write.
For creating new actions you could use bulk_create:https://docs.djangoproject.com/en/2.0/ref/models/querysets/#bulk-create
There is also this one: https://github.com/aykut/django-bulk-update (disclaimer: I am not a contributor or author of the package).
You have to be aware of cons of this method - if you use any post/pre_ save signals those will not be triggered by the update.
In general, running multiple saves will kill the database, and you might end up with hard to diagnose deadlocks. In one of the projects I worked on moving from save() in the loop into update() decreased response time from 30 something seconds to < 10 where the longest operations left where sending emails.

How to populate choice form from db in Django?

I can't figure out how to populate choice form from db. I know about ModelChoiceForm but the problem seems to be slightly different.
I want user to choose which sector does he work in. For example: 'Finance','Electronics' etc. which I would do simple:
SECTOR_CHOICES = (('finance',_('Finance'),
'electronics',_('Electronics')...
))
But the problem is that I want admin of the web to be able to add new choices, remove choice etc.
What came to my mind is to create a simple Model called Sector:
class Sector(models.Model):
name = models.CharField(max_length=40)
and User would have new attribute sector = models.ModelChoice(Sector).
But I'm scared what would happend when admin changes or removes a sector which is already used, and more, what if he removes it and the sector attribute is required?
How to solve this problem?

I would just override the delete_model as custom action and there check if the selected sector object is in use.
def delete_model(modeladmin, request, queryset):
for obj in queryset:
if UserModel.objects.filter(sector=obj).exists():
# do not delete, just add some message warning the admin about it
else:
obj.delete()
class UserModelAdmin(admin.ModelAdmin):
actions = [delete_model]
# ...

Inline-like solution for Django Admin where Admin contains ForeignKey to other model

I have several Customers who book Appointments. Each Appointment has exactly one customer, though a customer can be booked for multiple appointments occurring at different times.
class Customer(model.Model):
def __unicode__(self):
return u'%s' % (self.name,)
name = models.CharField(max_length=30)
# and about ten other fields I'd like to see from the admin view.
class Appointment(models.Model):
datetime = models.DateTimeField()
customer = models.ForeignKey("Customer")
class Meta:
ordering = ('datetime',)
Now when an admin goes to browse through the schedule by looking at the Appointments (ordered by time) in the admin, sometimes they want to see information about the customer who has a certain appointment. Right now, they'd have to remember the customer's name, navigate from the Appointment to the Customer admin page, find the remembered Customer, and only then could browse their information.
Ideally something like an admin inline would be great. However, I can only seem to make a CustomerInline on the Appointment admin page if Customer had a ForeignKey("Appointment"). (Django specifically gives me an error saying Customer has no ForeignKey to Appointment). Does anyone know of a similar functionality, but when Appointment has a ForeignKey('Customer')?
Note: I simplified the models; the actual Customer field currently has about ~10 fields besides the name (some free text), so it would be impractical to put all the information in the __unicode__.

There is no easy way to do this with django. The inlines are designed to follow relationships backwards.
Potentially the best substitute would be to provide a link to the user object. In the list view this is pretty trivial:
Add a method to your appointment model like:
def customer_admin_link(self):
return 'Customer' % reverse('admin:app_label_customer_change %s') % self.id
customer_admin_link.allow_tags = True
customer_admin_link.short_description = 'Customer'
Then in your ModelAdmin add:
list_display = (..., 'customer_admin_link', ...)
Another solution to get exactly what you're looking for at the cost of being a bit more complex would be to define a custom admin template. If you do that you can basically do anything. Here is a guide I've used before to explain:
http://www.unessa.net/en/hoyci/2006/12/custom-admin-templates/
Basically copy the change form from the django source and add code to display the customer information.

Completing #John's answer from above - define what you would like to see on the your changelist:
return '%s' % (
reverse('admin:applabel_customer_change', (self.customer.id,)),
self.customer.name # add more stuff here
)
And to add this to the change form, see: Add custom html between two model fields in Django admin's change_form

In the ModelAdmin class for your Appointments, you should declare the following method:
class MySuperModelAdmin(admin.ModelAdmin):
def get_form(self, request, obj=None, **kwargs):
if obj:
# create your own model admin instance here, because you will have the Customer's
# id so you know which instance to fetch
# something like the following
inline_instance = MyModelAdminInline(self.model, self.admin_site)
self.inline_instances = [inline_instance]
return super(MySuperModelAdmin, self).get_form(request, obj, **kwargs)
For more information, browser the source for that function to give you an idea of what you will have access to.
https://code.djangoproject.com/browser/django/trunk/django/contrib/admin/options.py#L423

There is a library you can use it.
https://github.com/daniyalzade/django_reverse_admin
But if you want to use link to object in showing table you can like this code:
def customer_link(self, obj):
if obj.customer:
reverse_link = 'admin:%s_%s_change' % (
obj.customer._meta.app_label, obj.customer._meta.model_name)
link = reverse(reverse_link, args=[obj.customer.id])
return format_html('More detail' % link)
return format_html('<span >-</span>')
customer_link.allow_tags = True
customer_link.short_description = 'Customer Info'
And in list_display:
list_display = (...,customer_link,...)