i need some help. I'm writting an app with django 1.6 and python 3.4.
My models:
class Account(models.Model):
number = models.IntegerField()
name = models.CharField(max_length = 100)
level = models.IntegerField()
parent_account = models.ForeignKey(Account, null=True, blank=True)
My View:
class InvoiceCreateView(CreateView):
model = Account
template_name = 'account/templates/create.html'
success_url = reverse_lazy('account_list')
Then when i create a new Account a dropdown list appear to select the Parent Account, everything fine, but i want to fill that dropdown list to accounts with level equal to 2 (example), not with all accounts.
something like this:
account.object.all().filter(level=2)
thanks for advance!
You can create your custom form and set the form_class variable in your view like:
class InvoiceCreateView(CreateView):
model = Account
form_class = AccountForm
template_name = 'account/templates/create.html'
success_url = reverse_lazy('account_list')
In your form, (I've named it AccountForm), you can define the parent field as desired:
class AccountForm(forms.ModelForm):
class Meta(object):
model = Account
parent = forms.ModelChoiceField(queryset=Account.objects.filter(level=2))
I guess this is what you are looking for:
account_list = account.objects.filter(level=2)
The Django Documentation on making queries also answers that question quite well.
Related
I am a newbie in django and I have a question about how I can save and show only the data of logged user - since my application is multi-tenant.
my view
class ProjetoCreate(CreateView):
model = Projeto
fields = ['nomeProjeto',
'descricao',
'dtInicio',
'deadline',
'nomeSprint',
'status',
]
def get_queryset(self):
logged_user = self.request.user
return Projeto.objects.filter(User=logged_user)
class ProjetoList(ListView):
paginate_by = 2
model = Projeto
my model
class Projeto(models.Model):
nomeProjeto = models.CharField(max_length=20)
descricao = HTMLField()
dtInicio = models.DateField(auto_now=False, auto_now_add=False)
deadline = models.DateField(auto_now=False, auto_now_add=False)
nomeSprint = models.CharField(max_length=30)
status = models.CharField(max_length=20)
Thank you very much!
Add
user = models.ForeignKey(User, on_delete=models.CASCADE)
to Projecto model. Then, in your view, set project.user = self.request.user before saving your project model.
I think you are doing it completely wrong.
You shouldn't be using get_queryset() at all in CreateView - https://stackoverflow.com/a/24043478/4626254
Here's is what you can try instead.
Add a user field in Project model and apply migrations.
user = models.ForeignKey(User, on_delete=models.CASCADE)
Create a class inheriting Generic APIView instead of CreateView.
Create a POST method like def post(self, request): inside that class and get all the details for creating a Projeto object in the request payload using request.data or request.POST.
Get the logged in user using request.user
Create a Projecto object with all this information as Projeto.objects.create(**your_other_fields, user=request.user)
Next time when filtering the objects, use a filter on user field like user=request.user.
I have a model with an attribute that is connected to another model as follow:
class Book(models.Model):
synced = models.OneToOneField('SyncedBook'
related_name='internal',
on_delete=models.CASCADE)
# some more attributes here...
#property
def book_address(self)
return self.synced.book_address
However, the book_address is a also a FK in the SyncedBook table as follow:
book_address = models.ForeignKey('Address', db_index=True, null=True, blank=True,
related_name='address_book', on_delete=models.PROTECT)
I don't know and understand how to be able to edit the book_address through the Django admin page in class BookingAdmin(admin.ModelAdmin), even though I have read over the documentation. At first I have the attribute as readonly, but now I want to be able to edit it and save the new address from the Address table. Is there a way to make it happen through the class BookingAdmin(admin.ModelAdmin) and how? Any example and solution would be appreciate
Model properties are typically used for presenting logically defined data for a particular model instance and not necessarily storing data on the model instance itself.
An example of when to use a model property is as follows:
# Defines a product instance
class Product(model.Models):
name = models.CharField(max_length=100)
description = models.TextField()
active = models.BooleanField(default=True)
cost = models.DecimalField(max_digits=5, decimal_places=2)
price = models.DecimalField(max_digits=5, decimal_places=2)
# calculate profits on product
#property
def profit(self)
p = self.price - self.cost
return p
In your case, you are trying to actually be able to modify data against a related model instance within the django admin. To me this sounds like more specifically an Inline (click here for documentation)
So in your case, you would need to create something like the following to your admin.py file:
class SyncedBookInline(admin.TabularInline):
model = BookInline
#admin.Register(Book)
class BookAdmin(admin.ModelAdmin):
# all your model admin settings
inlines = [SyncedBookInline]
Additional Info:
The Inline solution should still work for you. Please see the working code listed below:
models.py:
from django.db import models
class Hero(models.Model):
name = models.CharField(max_length=50)
class HeroAcquaintance(models.Model):
name = models.CharField(max_length=50)
hero = models.OneToOneField(Hero, on_delete=models.CASCADE)
admin.py:
from django.contrib import admin
from .models import *
class HeroAcquaintanceInline(admin.TabularInline):
model = HeroAcquaintance
#admin.register(Hero)
class HeroAdmin(admin.ModelAdmin):
list_display = (
'name',
)
inlines = [HeroAcquaintanceInline]
#admin.register(HeroAcquaintance)
class HeroAcquaintanceAdmin(admin.ModelAdmin):
list_display = (
'name',
)
Screenshot:
I have been getting my head around these basics but I am not getting it right. I am trying to associate my view to my user model using team which is a foreign key. When I try to create of a gps, I get an error saying "team is a required field" but instead it should be read only. The team attribute should be filled automatically with the id of the currentUser
Model
class User(models.Model):
first_name = models.CharField(max_length=200,blank=False)
last_name = models.CharField(max_length=200, blank=False)
class Gps(models.Model):
location = models.CharField(max_length=200,blank=False)
team= models.ForeignKey(User, on_delete=models.CASCADE)
serializers
class GpsSerializer(serializers.ModelSerializer):
class Meta:
model = Gps
fields = ('id','location','team')
view
class Gps_list(generics.ListCreateAPIView):
queryset = Gps.objects.all()
serializer_class = GpsSerializer
team = serializers.PrimaryKeyRelatedField(
read_only=True,
default=serializers.CurrentUserDefault()
)
There are two changes needed. First, team field definition should be moved to serializer class instead of view. Second, you should use Django's contrib.auth.User model instead of your definition of User, as because serializers.CurrentUserDefault() will bring request.user only. So you should remove your User definition and import that to your models.py:
from django.contrib.auth.models import User
Further steps would be to replace read_only=True with queryset=User.objects.all() to allow create.
I have a Django related question about foreign keys in the admin panel. I'm facing the following situation:
class Driver(models.Model):
name = models.CharField(max_length=200)
executable = models.CharField(max_length=200)
class Device(models.Model):
name = models.CharField(max_length=200)
bound_driver = models.ForeignKey(Driver)
class DriverAssignment(models.Model):
device = models.ForeignKey(Device)
driver = models.ForeignKey(Driver)
Every device needs to have a bound driver (which it uses). DriverAssignment should be the table which shows which driver can be used by which device. So one device can have multiple possibilities of drivers which can be bound. Now i would like to have a dropdown on my admin panel showing all possible drivers for a specific device to select the 'bound_driver'.
How can i do this in Django? This is probably an easy thing for an experienced Django guy. I hope someone can give me a hint since i'm kind of new to Django. Thanks a lot!
For Django >1.8
Use the InlineModelAdmin (docs for 2.2) as explained there:
models.py
from django.db import models
class Author(models.Model):
name = models.CharField(max_length=100)
class Book(models.Model):
author = models.ForeignKey(Author, on_delete=models.CASCADE)
title = models.CharField(max_length=100)
admin.py
from django.contrib import admin
class BookInline(admin.TabularInline):
model = Book
class AuthorAdmin(admin.ModelAdmin):
inlines = [
BookInline,
]
Change your model Structure to This:
class Driver(models.Model):
name = models.CharField(max_length=200)
executable = models.CharField(max_length=200)
class Device(models.Model):
name = models.CharField(max_length=200)
bound_driver = models.ForeignKey(Driver, related_name="bound_to")
available_drivers = models.ManyToManyfield(Driver)
ManyToManyField would do the same work as DriverAssignment Table.
You can add Available drivers in Available drivers field.
But then You would also Want that bound_driver is one of the Available Drivers. This validation you will have to do in forms. For that you have to over-ride Admin forms. See links
Links of Reference:
ManytoMany field: https://docs.djangoproject.com/en/1.6/ref/models/fields/#django.db.models.ManyToManyField
Model Admin (to over-ride admin functionality):
https://docs.djangoproject.com/en/1.6/ref/contrib/admin/#modeladmin-objects
You will have to spend some time reading and implementing if you want ot learn more. :)
OR
If you want to go with the same structure, than you will have to over-ride the form in ModelAdmin see here and Provide you custom form, which will be something like this:
class CustomForm(ModelForm)
bound_driver = forms.ModelChoiceField(queryset = <your custom queryset that returns only available drivers>, ...)
class Meta:
model = Device
https://docs.djangoproject.com/en/1.6/ref/contrib/admin/#django.contrib.admin.ModelAdmin.form
There is a snippet for inverse inlines. If you still need it you may try this:
https://gist.github.com/mzbyszewska/8b6afc312b024832aa85
It has been used by me for OneToOneField in django 1.5 and 1.6. Unfortunately I did not test it for ForeignKeyField, but the one of the previous users claims that it works for ForeignKeyField either.
The best description of the snippet is contained in it. The Person class is your DriverAssignment class and Device correspond to the Address class in the example below:
Example:
from django.db import models
class Address(models.Model):
street = models.CharField(max_length = 255)
zipcode = models.CharField(max_length = 10)
city = models.CharField(max_length = 255)
class Person(models.Model):
name = models.CharField(max_length = 255)
business_addr = models.ForeignKey(Address,
related_name = 'business_addr')
home_addr = models.OneToOneField(Address, related_name = 'home_addr')
other_addr = models.OneToOneField(Address, related_name = 'other_addr')
You use reverseadmin in the following way:
from django.contrib import admin
from django.db import models
from models import Person
from reverseadmin import ReverseModelAdmin
class AddressForm(models.Form):
pass
class PersonAdmin(ReverseModelAdmin):
inline_type = 'tabular'
inline_reverse = ('business_addr', ('home_addr', AddressForm), ('other_addr' (
'form': OtherForm
'exclude': ()
)))
admin.site.register(Person, PersonAdmin)
inline_type can be either "tabular" or "stacked" for tabular and
stacked inlines respectively.
I'm having trouble overriding the formset on a TabularInline inline of a ModelAdmin object in my admin site. I know you're supposed to have a model associated with a TabularInline object, but I'm not sure how to specify this on the form object used to generate the formset. With the code below, I'm getting "'AppAssetInline.formset' does not inherit from BaseModelFormSet."
class AppAssetForm(forms.ModelForm):
model = App.assets.through
primary = forms.BooleanField()
uuid = forms.CharField()
class AppAssetInline(admin.TabularInline):
model = App.assets.through
AssetFormset = formset_factory(AppAssetForm)
formset = AssetFormset
class AppAdmin(admin.ModelAdmin):
inlines = [AppAssetInline,]
The answer to my question didn't have to do with how I was structuring my forms, but rather how I was joining fields on my models. I had the following structure in my models:
class App(models.Model):
package = models.FileField(upload_to=settings.APP_PACKAGE_ROOT)
assets = models.ManyToManyField('AppAsset', blank=True, null=True)
download_count = models.IntegerField(default=0)
class AppAsset(models.Model):
def __unicode__(self):
return self.asset_file.name
notes = models.CharField(max_length=255, null=True, blank=True)
type = models.CharField(max_length=255, null=True, blank=True)
asset_file = models.FileField(upload_to=settings.APP_PACKAGE_ROOT)
What I did was change the structure such that AppAsset now has a foreign key on App for its assets. After that, I could use the TabularInline on the AppAsset model with no problems. Here are the latest source files:
https://github.com/ridecharge/spout/blob/master/Spout/AppDistribution/models.py
https://github.com/ridecharge/spout/blob/master/Spout/AppDistribution/admin.py
You should use django.forms.models.inlineformset_factory instead of formset_factory