Let me start by saying that I am working with a legacy database so avoiding the custom intermediate table is not an option.
I'm looking for an alternative way to get the limit_choices_to functionality as I need to only present the options flagged by the sample_option boolean in the Sampletype Model in my ModelForm:
class PlanetForm(ModelForm):
class Meta:
model = Planet
fields = ['name', 'samples']
Here is a simplified view of my models
class Planet(models.Model):
name= models.CharField(unique=True, max_length=256)
samples = models.ManyToManyField('Sampletype', through='Sample')
class Sample(models.Model):
planet = models.ForeignKey(Planet, models.DO_NOTHING)
sampletype = models.ForeignKey('Sampletype', models.DO_NOTHING)
class Sampletype(models.Model):
name = models.CharField(unique=True, max_length=256)
sample_option = models.BooleanField(default=True)
Sample is the intermediate table.
Normally, if the project had been started with Django in the first place, I could just define the ManyToManyField declaration as:
samples = models.ManyToManyField('Sampletype', limit_choices_to={'sample_option'=True})
But this is not an option.. So how do I get this functionality ?
Django clearly states in their documentation that:
limit_choices_to has no effect when used on a ManyToManyField with a
custom intermediate table specified using the through parameter.
But they offer no information on how to get that limit in place when you DO have a custom intermediate table.
I tried setting the limit_choices_to option on the ForeignKey in the Sample Model like so:
sampletype = models.ForeignKey('Sampletype', models.DO_NOTHING, limit_choices_to={'sample_option': True})
but that had no effect.
Strangely, I find no answer to this on the web and clearly other people must have to do this in their projects so I'm guessing the solution is really simple but I cannot figure it out.
Thanks in advance for any help or suggestions.
You could set the choices in the __init__ method of the form:
class PlanetForm(ModelForm):
class Meta:
model = Planet
fields = ['name', 'samples']
def __init__(self, *args, **kwargs):
super(PlanetForm, self).__init__(*args, **kwargs)
sample_choices = list(
Sampletype.objects.filter(sample_option=True).values_list('id', 'name')
)
# set these choices on the 'samples' field.
self.fields['samples'].choices = sample_choices
I've hit a dead end in my database-model design and could use some help.
We are using a Postgres database and django framework to store a variety of papers, books etc.
These are the base classes I'm having trouble with:
class Publisher(models.Model):
name = models.CharField(unique=True)
website = models.URLField(blank=True)
#...
class Editor(models.Model):
firstname = models.CharField()
lastname = models.CharField()
#...
class Book(models.Model):
title = models.CharField(unique=True)
#...
editors = models.ManyToManyField(Editor, blank=True)
publisher = models.ForeignKey(Publisher, blank=True)
#...
The Editor class above is modeled to store the editor(s) as persons of a corresponding book eg. "Ryan Carey" or "Nicholas Rowe".
The Publisher class would hold information like "University of Chicago Press".
The problem hatches should the Editor and Publisher be the same.
For example should the Editor and Publisher both be "University of Chicago Press".
Now, whenever a new Book is saved to the database through the django-admin interface I would like following to happen:
If one or more Editors are given, save it as in the model above.
If no Editors are given but a Publisher is given then make editors point to the same key publisher is pointing to
If none of both are given leave both blank
I was thinking of implementing it somehow like this:
class Book:
#...
def save(self, *args, **kwargs):
if self.editors.count() <= 0 and self.publisher:
self.editors = self.publisher #This is not possible
super(Book, self).save(*args, **kwargs)
The self.editors = self.publisher bit could perhaps be fixed with inheritance, but there are still multiple editors and just one publisher and I dont want publishers and editors to be stored int he same table.
Any thoughts on how to approach this?
With little of re-structuring your models, is possible and will be more flexible than now.
First: don't put publisher and editor to 2 separate models. Make one.
Second: if you can have publisher/editor both person and organization/company (and that will require different model fields), put all common fields into one model and make 2 models that will inherit from that model, containing more specified fields. That will create under the hood one-to-one relation between fields.
Third: create 2 fields in your book model, one ForeignKey named publisher and one ManyToMany named editors. Both relations should point to your base model for publisher/editor.
Sample code:
class PEBase(models.Model): # have no idea how to name it better
some_common_field = models.Field(...)
class PEPerson(PEBase):
first_name = models.CharField()
last_name = models.CharField()
class PEOrganization(PEBase):
name = models.CharField()
website = models.URLField()
class Book(models.Model):
title = models.CharField(unique=True)
#...
editors = models.ManyToManyField(PEBase, blank=True, related_name="books_edited")
publisher = models.ForeignKey(PEBase, blank=True, related_name="books_published")
def save(self, *args, **kwargs):
super(Book, self).save(*args, **kwargs)
if self.editors.count() <= 0 and self.publisher:
self.editors.add(self.publisher) #This must go after save - ID of book must be already set.
Example:
class MyUser(models.Model):
blocked_users = models.ManyToManyField("self", blank=True, null=True)
user = MyUser.object.get(pk=1)
user.blocked_users.add(user)
user.blocked_users.all()[0] == user # (!!!)
Can It be prevented on model/db level? Or we need just do check somewhere in app.
Looking at the Django docs for ManyToManyField arguments, it does not seem possible.
The closest argument to what you want is the limit_choices_to However, that only limits choices on ModelForms and admin (you can still save it like you did in your example), and there is currently no easy way to use it to limit based on another value (pk) in the current model.
If you want to prevent it from happening altogether, you'll have to resort to overriding the save method on the through model--something like:
class MyUser(models.Model):
blocked_users = models.ManyToManyField(..., through="BlockedUser")
class BlockedUser(models.Model):
user = models.ForeignKey(MyUser)
blocked = models.ForeignKey(MyUser)
def save(self, *args, **kwargs):
# Only allow this relationship to be created if
if self.user != self.blocked:
super(BlockedUser, self).save(*args, **kwargs)
You could of course also do this with signals.
I have the following (simplified) data structure:
Site
-> Zone
-> Room
-> name
I want the name of each Room to be unique for each Site.
I know that if I just wanted uniqueness for each Zone, I could do:
class Room(models.Model):
zone = models.ForeignKey(Zone)
name = models.CharField(max_length=255)
class Meta:
unique_together = ('name', 'zone')
But I can't do what I really want, which is:
class Room(models.Model):
zone = models.ForeignKey(Zone)
name = models.CharField(max_length=255)
class Meta:
unique_together = ('name', 'zone__site')
I tried adding a validate_unique method, as suggested by this question:
class Room(models.Model):
zone = models.ForeignKey(Zone)
name = models.CharField(max_length=255)
def validate_unique(self, exclude=None):
qs = Room.objects.filter(name=self.name)
if qs.filter(zone__site=self.zone__site).exists():
raise ValidationError('Name must be unique per site')
models.Model.validate_unique(self, exclude=exclude)
but I must be misunderstanding the point/implementation of validate_unique, because it is not being called when I save a Room object.
What would be the correct way to implement this check?
Methods are not called on their own when saving the model.
One way to do this is to have a custom save method that calls the validate_unique method when a model is saved:
class Room(models.Model):
zone = models.ForeignKey(Zone)
name = models.CharField(max_length=255)
def validate_unique(self, exclude=None):
qs = Room.objects.filter(name=self.name)
if qs.filter(zone__site=self.zone__site).exists():
raise ValidationError('Name must be unique per site')
def save(self, *args, **kwargs):
self.validate_unique()
super(Room, self).save(*args, **kwargs)
class Room(models.Model):
zone = models.ForeignKey(Zone)
name = models.CharField(max_length=255)
def validate_unique(self, *args, **kwargs):
super(Room, self).validate_unique(*args, **kwargs)
qs = Room.objects.filter(name=self.name)
if qs.filter(zone__site=self.zone__site).exists():
raise ValidationError({'name':['Name must be unique per site',]})
I needed to make similar program. It worked.
The Django Validation objects documentation explains the steps involved in validation including this snippet
Note that full_clean() will not be called automatically when you call your model's save() method
If the model instance is being created as a result of using a ModelForm, then validation will occur when the form is validated.
There are a some options in how you handle validation.
Call the model instance's full_clean() manually before saving.
Override the save() method of the model to perform validation on every save. You can choose how much validation should occur here, whether you want full validation or only uniqueness checks.
class Room(models.Model):
def save(self, *args, **kwargs):
self.full_clean()
super(Room, self).save(*args, **kwargs)
Use a Django pre_save signal handler which will automatically perform validation before a save. This provides a very simple way to add validation on exisiting models without any additional model code.
# In your models.py
from django.db.models.signals import pre_save
def validate_model_signal_handler(sender, **kwargs):
"""
Signal handler to validate a model before it is saved to database.
"""
# Ignore raw saves.
if not kwargs.get('raw', False):
kwargs['instance'].full_clean()
pre_save.connect(validate_model_signal_handler,
sender=Room,
dispatch_uid='validate_model_room')
I have a two way foreign relation similar to the following
class Parent(models.Model):
name = models.CharField(max_length=255)
favoritechild = models.ForeignKey("Child", blank=True, null=True)
class Child(models.Model):
name = models.CharField(max_length=255)
myparent = models.ForeignKey(Parent)
How do I restrict the choices for Parent.favoritechild to only children whose parent is itself? I tried
class Parent(models.Model):
name = models.CharField(max_length=255)
favoritechild = models.ForeignKey("Child", blank=True, null=True, limit_choices_to = {"myparent": "self"})
but that causes the admin interface to not list any children.
I just came across ForeignKey.limit_choices_to in the Django docs.
Not sure yet how it works, but it might be the right thing here.
Update: ForeignKey.limit_choices_to allows one to specify either a constant, a callable or a Q object to restrict the allowable choices for the key. A constant obviously is of no use here, since it knows nothing about the objects involved.
Using a callable (function or class method or any callable object) seems more promising. However, the problem of how to access the necessary information from the HttpRequest object remains. Using thread local storage may be a solution.
2. Update: Here is what has worked for me:
I created a middleware as described in the link above. It extracts one or more arguments from the request's GET part, such as "product=1", and stores this information in the thread locals.
Next there is a class method in the model that reads the thread local variable and returns a list of ids to limit the choice of a foreign key field.
#classmethod
def _product_list(cls):
"""
return a list containing the one product_id contained in the request URL,
or a query containing all valid product_ids if not id present in URL
used to limit the choice of foreign key object to those related to the current product
"""
id = threadlocals.get_current_product()
if id is not None:
return [id]
else:
return Product.objects.all().values('pk').query
It is important to return a query containing all possible ids if none was selected so that the normal admin pages work ok.
The foreign key field is then declared as:
product = models.ForeignKey(
Product,
limit_choices_to={
id__in=BaseModel._product_list,
},
)
The catch is that you have to provide the information to restrict the choices via the request. I don't see a way to access "self" here.
The 'right' way to do it is to use a custom form. From there, you can access self.instance, which is the current object. Example --
from django import forms
from django.contrib import admin
from models import *
class SupplierAdminForm(forms.ModelForm):
class Meta:
model = Supplier
fields = "__all__" # for Django 1.8+
def __init__(self, *args, **kwargs):
super(SupplierAdminForm, self).__init__(*args, **kwargs)
if self.instance:
self.fields['cat'].queryset = Cat.objects.filter(supplier=self.instance)
class SupplierAdmin(admin.ModelAdmin):
form = SupplierAdminForm
The new "right" way of doing this, at least since Django 1.1 is by overriding the AdminModel.formfield_for_foreignkey(self, db_field, request, **kwargs).
See http://docs.djangoproject.com/en/1.2/ref/contrib/admin/#django.contrib.admin.ModelAdmin.formfield_for_foreignkey
For those who don't want to follow the link below is an example function that is close for the above questions models.
class MyModelAdmin(admin.ModelAdmin):
def formfield_for_foreignkey(self, db_field, request, **kwargs):
if db_field.name == "favoritechild":
kwargs["queryset"] = Child.objects.filter(myparent=request.object_id)
return super(MyModelAdmin, self).formfield_for_manytomany(db_field, request, **kwargs)
I'm only not sure about how to get the current object that is being edited. I expect it is actually on the self somewhere but I'm not sure.
This isn't how django works. You would only create the relation going one way.
class Parent(models.Model):
name = models.CharField(max_length=255)
class Child(models.Model):
name = models.CharField(max_length=255)
myparent = models.ForeignKey(Parent)
And if you were trying to access the children from the parent you would do
parent_object.child_set.all(). If you set a related_name in the myparent field, then that is what you would refer to it as. Ex: related_name='children', then you would do parent_object.children.all()
Read the docs http://docs.djangoproject.com/en/dev/topics/db/models/#many-to-one-relationships for more.
If you only need the limitations in the Django admin interface, this might work. I based it on this answer from another forum - although it's for ManyToMany relationships, you should be able to replace formfield_for_foreignkey for it to work. In admin.py:
class ParentAdmin(admin.ModelAdmin):
def get_form(self, request, obj=None, **kwargs):
self.instance = obj
return super(ParentAdmin, self).get_form(request, obj=obj, **kwargs)
def formfield_for_foreignkey(self, db_field, request=None, **kwargs):
if db_field.name == 'favoritechild' and self.instance:
kwargs['queryset'] = Child.objects.filter(myparent=self.instance.pk)
return super(ChildAdmin, self).formfield_for_foreignkey(db_field, request=request, **kwargs)
#Ber: I have added validation to the model similar to this
class Parent(models.Model):
name = models.CharField(max_length=255)
favoritechild = models.ForeignKey("Child", blank=True, null=True)
def save(self, force_insert=False, force_update=False):
if self.favoritechild is not None and self.favoritechild.myparent.id != self.id:
raise Exception("You must select one of your own children as your favorite")
super(Parent, self).save(force_insert, force_update)
which works exactly how I want, but it would be really nice if this validation could restrict choices in the dropdown in the admin interface rather than validating after the choice.
I'm trying to do something similar. It seems like everyone saying 'you should only have a foreign key one way' has maybe misunderstood what you're trying do.
It's a shame the limit_choices_to={"myparent": "self"} you wanted to do doesn't work... that would have been clean and simple. Unfortunately the 'self' doesn't get evaluated and goes through as a plain string.
I thought maybe I could do:
class MyModel(models.Model):
def _get_self_pk(self):
return self.pk
favourite = models.ForeignKey(limit_choices_to={'myparent__pk':_get_self_pk})
But alas that gives an error because the function doesn't get passed a self arg :(
It seems like the only way is to put the logic into all the forms that use this model (ie pass a queryset in to the choices for your formfield). Which is easily done, but it'd be more DRY to have this at the model level. Your overriding the save method of the model seems a good way to prevent invalid choices getting through.
Update
See my later answer for another way https://stackoverflow.com/a/3753916/202168
Do you want to restrict the choices available in the admin interface when creating/editing a model instance?
One way to do this is validation of the model. This lets you raise an error in the admin interface if the foreign field is not the right choice.
Of course, Eric's answer is correct: You only really need one foreign key, from child to parent here.
An alternative approach would be not to have 'favouritechild' fk as a field on the Parent model.
Instead you could have an is_favourite boolean field on the Child.
This may help:
https://github.com/anentropic/django-exclusivebooleanfield
That way you'd sidestep the whole problem of ensuring Children could only be made the favourite of the Parent they belong to.
The view code would be slightly different but the filtering logic would be straightforward.
In the admin you could even have an inline for Child models that exposed the is_favourite checkbox (if you only have a few children per parent) otherwise the admin would have to be done from the Child's side.
A much simpler variation of #s29's answer:
Instead of customising the form,
You can simply restrict the choices available in form field from your view:
what worked for me was:
in forms.py:
class AddIncomingPaymentForm(forms.ModelForm):
class Meta:
model = IncomingPayment
fields = ('description', 'amount', 'income_source', 'income_category', 'bank_account')
in views.py:
def addIncomingPayment(request):
form = AddIncomingPaymentForm()
form.fields['bank_account'].queryset = BankAccount.objects.filter(profile=request.user.profile)
from django.contrib import admin
from sopin.menus.models import Restaurant, DishType
class ObjInline(admin.TabularInline):
def __init__(self, parent_model, admin_site, obj=None):
self.obj = obj
super(ObjInline, self).__init__(parent_model, admin_site)
class ObjAdmin(admin.ModelAdmin):
def get_inline_instances(self, request, obj=None):
inline_instances = []
for inline_class in self.inlines:
inline = inline_class(self.model, self.admin_site, obj)
if request:
if not (inline.has_add_permission(request) or
inline.has_change_permission(request, obj) or
inline.has_delete_permission(request, obj)):
continue
if not inline.has_add_permission(request):
inline.max_num = 0
inline_instances.append(inline)
return inline_instances
class DishTypeInline(ObjInline):
model = DishType
def formfield_for_foreignkey(self, db_field, request=None, **kwargs):
field = super(DishTypeInline, self).formfield_for_foreignkey(db_field, request, **kwargs)
if db_field.name == 'dishtype':
if self.obj is not None:
field.queryset = field.queryset.filter(restaurant__exact = self.obj)
else:
field.queryset = field.queryset.none()
return field
class RestaurantAdmin(ObjAdmin):
inlines = [
DishTypeInline
]