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Add custom form fields that are not part of the model (Django)

I have a model registered on the admin site. One of its fields is a long string expression. I'd like to add custom form fields to the add/update pages of this model in the admin. Based on the values of these fields I will build the long string expression and save it in the relevant model field.
How can I do this?
I'm building a mathematical or string expression from symbols. The user chooses symbols (these are the custom fields that are not part of the model) and when they click save then I create a string expression representation from the list of symbols and store it in the DB. I don't want the symbols to be part of the model and DB, only the final expression.

Either in your admin.py or in a separate forms.py you can add a ModelForm class and then declare your extra fields inside that as you normally would. I've also given an example of how you might use these values in form.save():
from django import forms
from yourapp.models import YourModel
class YourModelForm(forms.ModelForm):
extra_field = forms.CharField()
def save(self, commit=True):
extra_field = self.cleaned_data.get('extra_field', None)
# ...do something with extra_field here...
return super(YourModelForm, self).save(commit=commit)
class Meta:
model = YourModel
To have the extra fields appearing in the admin just:
Edit your admin.py and set the form property to refer to the form you created above.
Include your new fields in your fields or fieldsets declaration.
Like this:
class YourModelAdmin(admin.ModelAdmin):
form = YourModelForm
fieldsets = (
(None, {
'fields': ('name', 'description', 'extra_field',),
}),
)
UPDATE:
In Django 1.8 you need to add fields = '__all__' to the metaclass of YourModelForm.

It it possible to do in the admin, but there is not a very straightforward way to it. Also, I would like to advice to keep most business logic in your models, so you won't be dependent on the Django Admin.
Maybe it would be easier (and maybe even better) if you have the two seperate fields on your model. Then add a method on your model that combines them.
For example:
class MyModel(models.model):
field1 = models.CharField(max_length=10)
field2 = models.CharField(max_length=10)
def combined_fields(self):
return '{} {}'.format(self.field1, self.field2)
Then in the admin you can add the combined_fields() as a readonly field:
class MyModelAdmin(models.ModelAdmin):
list_display = ('field1', 'field2', 'combined_fields')
readonly_fields = ('combined_fields',)
def combined_fields(self, obj):
return obj.combined_fields()
If you want to store the combined_fields in the database you could also save it when you save the model:
def save(self, *args, **kwargs):
self.field3 = self.combined_fields()
super(MyModel, self).save(*args, **kwargs)

Django 2.1.1
The primary answer got me halfway to answering my question. It did not help me save the result to a field in my actual model. In my case I wanted a textfield that a user could enter data into, then when a save occurred the data would be processed and the result put into a field in the model and saved. While the original answer showed how to get the value from the extra field, it did not show how to save it back to the model at least in Django 2.1.1
This takes the value from an unbound custom field, processes, and saves it into my real description field:
class WidgetForm(forms.ModelForm):
extra_field = forms.CharField(required=False)
def processData(self, input):
# example of error handling
if False:
raise forms.ValidationError('Processing failed!')
return input + " has been processed"
def save(self, commit=True):
extra_field = self.cleaned_data.get('extra_field', None)
# self.description = "my result" note that this does not work
# Get the form instance so I can write to its fields
instance = super(WidgetForm, self).save(commit=commit)
# this writes the processed data to the description field
instance.description = self.processData(extra_field)
if commit:
instance.save()
return instance
class Meta:
model = Widget
fields = "__all__"

You can always create new admin template, and do what you need in your admin_view (override the admin add URL to your admin_view):
url(r'^admin/mymodel/mymodel/add/$','admin_views.add_my_special_model')

If you absolutely only want to store the combined field on the model and not the two seperate fields, you could do something like this:
Create a custom form using the form attribute on your ModelAdmin. ModelAdmin.form
Parse the custom fields in the save_formset method on your ModelAdmin. ModelAdmin.save_model(request, obj, form, change)
I never done something like this so I'm not completely sure how it will work out.

The first (highest score) solution (https://stackoverflow.com/a/23337009/10843740) was accurate, but I have more.
If you declare fields by code, that solution works perfectly, but what if you want to build those dynamically?
In this case, creating fields in the __init__ function for the ModelForm won't work. You will need to pass a custom metaclass and override the declared_fields in the __new__ function!
Here is a sample:
class YourCustomMetaClass(forms.models.ModelFormMetaclass):
"""
For dynamically creating fields in ModelForm to be shown on the admin panel,
you must override the `declared_fields` property of the metaclass.
"""
def __new__(mcs, name, bases, attrs):
new_class = super(NamedTimingMetaClass, mcs).__new__(
mcs, name, bases, attrs)
# Adding fields dynamically.
new_class.declared_fields.update(...)
return new_class
# don't forget to pass the metaclass
class YourModelForm(forms.ModelForm, metaclass=YourCustomMetaClass):
"""
`metaclass=YourCustomMetaClass` is where the magic happens!
"""
# delcare static fields here
class Meta:
model = YourModel
fields = '__all__'

This is what I did to add the custom form field "extra_field" which is not the part of the model "MyModel" as shown below:
# "admin.py"
from django.contrib import admin
from django import forms
from .models import MyModel
class MyModelForm(forms.ModelForm):
extra_field = forms.CharField()
def save(self, commit=True):
extra_field = self.cleaned_data.get('extra_field', None)
# Do something with extra_field here
return super().save(commit=commit)
#admin.register(MyModel)
class MyModelAdmin(admin.ModelAdmin):
form = MyModelForm

You might get help from my answer at :
my response previous on multicheckchoice custom field
You can also extend multiple forms having different custom fields and then assigning them to your inlines class like stackedinline or tabularinline:
form =
This way you can avoid formset complication where you need to add multiple custom fields from multiple models.
so your modeladmin looks like:
inlines = [form1inline, form2inline,...]
In my previous response to the link here, you will find init and save methods.
init will load when you view the page and save will send it to database.
in these two methods you can do your logic to add strings and then save thereafter view it back in Django admin change_form or change_list depending where you want.
list_display will show your fields on change_list.
Let me know if it helps ...
....
class CohortDetailInline3(admin.StackedInline):
model = CohortDetails
form = DisabilityTypesForm
...
class CohortDetailInline2(admin.StackedInline):
model = CohortDetails
form = StudentRPLForm
...
...
#admin.register(Cohort)
class CohortAdmin(admin.ModelAdmin):
form = CityInlineForm
inlines = [uploadInline, cohortDetailInline1,
CohortDetailInline2, CohortDetailInline3]
list_select_related = True
list_display = ['rto_student_code', 'first_name', 'family_name',]
...

Adding Custom Django Model Validation

I have a Django model with a start and end date range. I want to enforce validation so that no two records have overlapping date ranges. What's the simplest way to implement this so that I don't have to repeat myself writing this logic?
e.g. I don't want to re-implement this logic in a Form and a ModelForm and an admin form and the model's overridden save().
As far as I know, Django doesn't make it easy to globally enforce these types of criteria.
Googling hasn't been very helpful, since "model validation" typically refers to validating specific model fields, and not the entire model contents, or relations between fields.

The basic pattern I've found useful is to put all my custom validation in clean() and then simply call full_clean() (which calls clean() and a few other methods) from inside save(), e.g.:
class BaseModel(models.Model):
def clean(self, *args, **kwargs):
# add custom validation here
super().clean(*args, **kwargs)
def save(self, *args, **kwargs):
self.full_clean()
super().save(*args, **kwargs)
This isn't done by default, as explained here, because it interferes with certain features, but those aren't a problem for my application.

I would override the validate_unique method on the model. To make sure you ignore the current object when validating, you can use the following:
from django.db.models import Model, DateTimeField
from django.core.validators import NON_FIELD_ERRORS, ValidationError
class MyModel(Model):
start_date = DateTimeField()
end_date = DateTimeField()
def validate_unique(self, *args, **kwargs):
super(MyModel, self).validate_unique(*args, **kwargs)
qs = self.__class__._default_manager.filter(
start_date__lt=self.end_date,
end_date__gt=self.start_date
)
if not self._state.adding and self.pk is not None:
qs = qs.exclude(pk=self.pk)
if qs.exists():
raise ValidationError({
NON_FIELD_ERRORS: ['overlapping date range',],
})
ModelForm will automatically call this for you through a full_clean(), which you can use manually too.
PPR has a nice discussion of a simple, correct range overlap condition.

I think you should use this:
https://docs.djangoproject.com/en/dev/ref/models/instances/#validating-objects
Just define clean() method in your model like this: (example from the docs link)
def clean(self):
from django.core.exceptions import ValidationError
# Don't allow draft entries to have a pub_date.
if self.status == 'draft' and self.pub_date is not None:
raise ValidationError('Draft entries may not have a publication date.')
# Set the pub_date for published items if it hasn't been set already.
if self.status == 'published' and self.pub_date is None:
self.pub_date = datetime.datetime.now()

I think this can help you,
We can create multiple validators like this use in models.
from django.core.exceptions import ValidationError
from django.utils.translation import gettext_lazy as _
from django.db import models
def validate_even(value):
if value % 2 != 0:
raise ValidationError(
_('%(value)s is not an even number'),
params={'value': value},
)
class MyModel(models.Model):
even_field = models.IntegerField(validators=[validate_even])

Populating django field with pre_save()?

class TodoList(models.Model):
title = models.CharField(maxlength=100)
slug = models.SlugField(maxlength=100)
def save(self):
self.slug = title
super(TodoList, self).save()
I'm assuming the above is how to create and store a slug when a title is inserted into the table TodoList, if not, please correct me!
Anyhow, I've been looking into pre_save() as another way to do this, but can't figure out how it works. How do you do it with pre_save()?
Is it like the below code snippet?
def pre_save(self):
self.slug = title
I'm guessing not. What is the code to do this?
Thanks!

Most likely you are referring to django's pre_save signal. You could setup something like this:
from django.db.models.signals import pre_save
from django.dispatch import receiver
from django.template.defaultfilters import slugify
#receiver(pre_save)
def my_callback(sender, instance, *args, **kwargs):
instance.slug = slugify(instance.title)
If you dont include the sender argument in the decorator, like #receiver(pre_save, sender=MyModel), the callback will be called for all models.
You can put the code in any file that is parsed during the execution of your app, models.py is a good place for that.

#receiver(pre_save, sender=TodoList)
def my_callback(sender, instance, *args, **kwargs):
instance.slug = slugify(instance.title)

you can use django signals.pre_save:
from django.db.models.signals import post_save, post_delete, pre_save
class TodoList(models.Model):
#staticmethod
def pre_save(sender, instance, **kwargs):
#do anything you want
pre_save.connect(TodoList.pre_save, TodoList, dispatch_uid="sightera.yourpackage.models.TodoList")

The pre_save() signal hook is indeed a great place to handle slugification for a large number of models. The trick is to know what models need slugs generated, what field should be the basis for the slug value.
I use a class decorator for this, one that lets me mark models for auto-slug-generation, and what field to base it on:
from django.db import models
from django.dispatch import receiver
from django.utils.text import slugify
def autoslug(fieldname):
def decorator(model):
# some sanity checks first
assert hasattr(model, fieldname), f"Model has no field {fieldname!r}"
assert hasattr(model, "slug"), "Model is missing a slug field"
#receiver(models.signals.pre_save, sender=model, weak=False)
def generate_slug(sender, instance, *args, raw=False, **kwargs):
if not raw and not instance.slug:
source = getattr(instance, fieldname)
slug = slugify(source)
if slug: # not all strings result in a slug value
instance.slug = slug
return model
return decorator
This registers a signal handler for specific models only, and lets you vary the source field with each model decorated:
#autoslug("name")
class NamedModel(models.Model):
name = models.CharField(max_length=100)
slug = models.SlugField()
#autoslug("title")
class TitledModel(models.Model):
title = models.CharField(max_length=255)
slug = models.SlugField()
Note that no attempt is made to generate a unique slug value. That would require checking for integrity exceptions in a transaction or using a randomised value in the slug from a large enough pool as to make collisions unlikely. Integrity exception checking can only be done in the save() method, not in signal hooks.

Receiver functions must be like this:
def my_callback(sender, **kwargs):
print("Request finished!")
Notice that the function takes a sender argument, along with wildcard keyword arguments (**kwargs); all signal handlers must take these arguments.
All signals send keyword arguments, and may change those keyword arguments at any time.
Reference here.

Dynamically setting the queryset of a ModelMultipleChoiceField to a custom recordset

I've seen all the howtos about how you can set a ModelMultipleChoiceField to use a custom queryset and I've tried them and they work. However, they all use the same paradigm: the queryset is just a filtered list of the same objects.
In my case, I'm trying to get the admin to draw a multiselect form that instead of using usernames as the text portion of the , I'd like to use the name field from my account class.
Here's a breakdown of what I've got:
# models.py
class Account(models.Model):
name = models.CharField(max_length=128,help_text="A display name that people understand")
user = models.ForeignKey(User, unique=True) # Tied to the User class in settings.py
class Organisation(models.Model):
administrators = models.ManyToManyField(User)
# admin.py
from django.forms import ModelMultipleChoiceField
from django.contrib.auth.models import User
class OrganisationAdminForm(forms.ModelForm):
def __init__(self, *args, **kwargs):
from ethico.accounts.models import Account
self.base_fields["administrators"] = ModelMultipleChoiceField(
queryset=User.objects.all(),
required=False
)
super(OrganisationAdminForm, self).__init__(*args, **kwargs)
class Meta:
model = Organisation
This works, however, I want queryset above to draw a selectbox with the Account.name property and the User.id property. This didn't work:
queryset=Account.objects.all().order_by("name").values_list("user","name")
It failed with this error:
'tuple' object has no attribute 'pk'
I figured that this would be easy, but it's turned into hours of dead-ends. Anyone care to shed some light?

You can use a custom widget, override its render method. Here's what I had done for a text field :
class UserToAccount(forms.widgets.TextInput):
def render(self, name, value, attrs=None):
if isinstance(value, User) :
value = Account.objects.get(user=value).name
return super (UserToAccount, self).render(name, value, attrs=None)
Then of course, use the widget parameter of your administrator field, in order to use your custom widget.
I don't know if it can be adapted for a select, but you can try out.

The queryset needs to be a QuerySet, when you do values_list you get a list so that won't work.
If you want to change the default display of models, just override __unicode__. See http://docs.djangoproject.com/en/dev/ref/models/instances/#unicode
For example:
def __unicode__(self):
return u"%s for %s" % (self.name, self.user)
Django will use __unicode__ whenever you asks it to print a model. For testing you can just load up a model in the shell and do print my_instance.

Taking a queue from sebpiq, I managed to figure it out:
class OrganisationAdminForm(forms.ModelForm):
def __init__(self, *args, **kwargs):
from django.forms import MultipleChoiceField
from ethico.accounts.models import Account
self.base_fields["administrators"] = MultipleChoiceField(
choices=tuple([(a.user_id, a.name) for a in Account.objects.all().order_by("name")]),
widget=forms.widgets.SelectMultiple,
required=False
)
super(OrganisationAdminForm, self).__init__(*args, **kwargs)
class Meta:
model = Organisation
class OrganisationAdmin(admin.ModelAdmin):
form = OrganisationAdminForm
admin.site.register(Organisation, OrganisationAdmin)
The key was abandoning the queryset altogether. Once I went with a fixed choices= parameter, everything just worked. Thanks everyone!

How do I restrict foreign keys choices to related objects only in django

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
]