I need to add a field validator through an abstract class.
The code I'm working with supposes that every class inheriting from this abstract class has a field name, but this field itself isn't defined in the abstract class unfortunately.
So I tried the following, but I'm not sure what is the good way of finding the field in self._meta.fields, since this is a list..??
class AbsClass(models.Model):
class Meta:
abstract = True
def __init__(self, *args, **kw):
super(AbsClass, self).__init__(*args, **kw)
name_field = [f for f in self._meta.fields if f.name == 'name'][0] # Is there another way?
name_field.validators.append(my_validator)
You need Options.get_field:
...
name_field = self._meta.get_field('name')
name_field.validators.append(my_validator)
...
Your approach, however, doesn't seem like a good idea: you model's field instances are shared between all instances of your model (their references are stored in a class attribute, not in instance attributes). This means that every time you instantiate an object of your model, you'll be adding another copy of my_validator to the field's validators, because you're adding it to the same field instance.
You could implement a metaclass for your abstract base class and add the validator at compile time instead of tampering with field instances at runtime, something along the lines of this (not tested):
from django.utils.six import with_metaclass
from django.db.models.base import ModelBase
# inherit from Django's model metaclass
class AbsClassMeta(ModelBase):
def __new__(cls, name, bases, attrs):
if 'name' in attrs:
attrs['name'].validators.append(my_validator)
elif name != 'AbsClass':
# is it an error to not have a "name" field in your subclasses?
# handle situation appropriately
return super(AbsClassMeta, cls).__new__(cls, name, bases, attrs)
class AbsClass(with_metaclass(AbsClassMeta, models.Model)):
...
Note that your AbsClass class itself will also be created using this metaclass. If you decide to throw an exception in AbsClassMeta.__new__ if the class doesn't have a name field, you need to take this into account, since your AbsClass class doesn't have a name field.
In Django 2.x
instead of
name_field = self._meta.get_field('name')
name_field.validators.append(my_validator)
you can do:
name_field = self.fields['name']
name_field.validators.append(my_validator)
Related
I am adding a slug to all my models for serialization purposes, so I have defined an abstract base class which uses the AutoSlugField from django_autoslug.
class SluggerModel(models.Model):
slug = AutoSlugField(unique=True, db_index=False)
class Meta:
abstract=True
I also have a custom manager and a natural_key method defined, and at this point I have about 20 child classes, so there are several things that make using an abstract base model worthwhile besides just the single line that defines the field.
However, I want to be able to switch a few of the default arguments for initializing the AutoSlugField for some of the child models, while still being able to utilize the abstract base class. For example, I'd like some of them to utilize the populate_from option, specifiying fields from their specific model, and others to have db_index=True instead of my default (False).
I started trying to do this with a custom Metaclass, utilizing custom options defined in each child Model's inner Meta class, but thats become a rat's nest. I'm open to guidance on that approach, or any other suggestions.
One solution would be to dynamically construct your abstract base class. For example:
def get_slugger_model(**slug_kwargs):
defaults = {
'unique': True,
'db_index': False
}
defaults.update(slug_kwargs)
class MySluggerModel(models.Model):
slug = AutoSlugField(**defaults)
class Meta:
abstract = True
return MySluggerModel
class MyModel(get_slugger_model()):
pass
class MyModel2(get_slugger_model(populate_from='name')):
name = models.CharField(max_length=20)
Update: I started out with the following solution, which was ugly, and switched to Daniel's solution, which is not. I'm leaving mine here for reference.
Here's my Metaclass rat trap that seems to be working (without extensive testing yet).
class SluggerMetaclass(ModelBase):
"""
Metaclass hack that provides for being able to define 'slug_from' and
'slug_db_index' in the Meta inner class of children of SluggerModel in order to set
those properties on the AutoSlugField
"""
def __new__(cls, name, bases, attrs):
# We don't want to add this to the SluggerModel class itself, only its children
if name != 'SluggerModel' and SluggerModel in bases:
_Meta = attrs.get('Meta', None)
if _Meta and hasattr(_Meta, 'slug_from') or hasattr(_Meta, 'slug_db_index'):
attrs['slug'] = AutoSlugField(
populate_from=getattr(_Meta, 'slug_from', None),
db_index=getattr(_Meta, 'slug_db_index', False),
unique=True
)
try:
# ModelBase will reject unknown stuff in Meta, so clear it out before calling super
delattr(_Meta, 'slug_from')
except AttributeError:
pass
try:
delattr(_Meta, 'slug_db_index')
except AttributeError:
pass
else:
attrs['slug'] = AutoSlugField(unique=True, db_index = False) # default
return super(SlugSerializableMetaclass, cls).__new__(cls, name, bases, attrs)
The SlugModel looks basically like this now:
class SluggerModel(models.Model):
__metaclass__ = SluggerMetaclass
objects = SluggerManager()
# I don't define the AutoSlugField here because the metaclass will add it to the child class.
class Meta:
abstract = True
And I can acheive the desired effect with:
class SomeModel(SluggerModel, BaseModel):
name = CharField(...)
class Meta:
slug_from = 'name'
slug_db_index = True
I have to put SluggerModel first in the inheritance list for models having more than one abstract parent model, or else the fields aren't picked up by the other parent models and validation fails; however, I couldn't decipher why.
I guess I could put this an answer to my own question, since it works, but I'm hoping for a better way since its a bit on the ugly side. Then again, hax is hax so what can you do, so maybe this is the answer.
This seems like a bug but I just want to make sure I'm consuming the API properly.
It seems that support for django's modelform isn't supported on neo4django. Here's what I have:
Simple class:
from neo4django.db import models
class Person(models.NodeModel):
name = models.StringProperty()
The modelform:
class PersonForm(forms.ModelForm):
class Meta:
model = Person
Will trigger exception:
'super' object has no attribute 'editable'
I posted details as an issue:
https://github.com/scholrly/neo4django/issues/135
Because when Django goes to lookup field information using the model's _meta information, it finds a BoundProperty instead of a StringProperty or Property (which has a member called 'editable', but BoundProperty doesn't).
Is there a workaround, or is this an actual bug? Any ideas on how to fix the bug? I'm not familiar with the library codebase.
Thanks!
Below is a reasonable (and quick) workaround for anyone using neo4j with Django.
This solution requires that field names on the form have the exact same name as the attributes of the model.
Inherit the form from this class and set the model under the form class Meta class:
class NeoModelForm(forms.Form):
def __init__(self, *args, **kwargs):
super(NeoModelForm, self).__init__(*args, **kwargs)
self._meta = getattr(self, 'Meta', None)
if not self._meta:
raise Exception('Missing Meta class on %s' % str(self.__class__.__name__))
if not hasattr(self._meta, 'model'):
raise Exception('Missing model on Meta class of %s' % str(self.__class__.__name__))
def save(self, commit=True):
if not self.is_valid():
raise Exception('Failed to validate')
instance = self._meta.model(**self.cleaned_data)
if commit:
instance.save()
return instance
Now you can create a form class like this:
class PersonForm(NeoModelForm):
name = forms.CharField(widget=forms.TextInput())
class Meta:
model = Person
And still be able to save a model instance from a valid form:
form = formclass(request.POST)
if form.is_valid():
obj = form.save()
Plus the commit argument will give you the same solution as django's modelform class- but I didn't bother to implement to save_m2m functionality (which doesn't seem relevant for neo4j as a backend).
I want my Django custom model field to set an attribute on the model instance.
I'm sure it's not working this way but here is an example:
class MyField(models.Field):
__metaclass__ = models.SubfieldBase
def __init__(self, *args, **kwargs):
super(MyField, self).__init__(*args, **kwargs)
model_instance = ????
setattr(model_instance, "extra_attribute", "It's working!")
class MyModel(models.Model):
my_field = MyField()
model_instance = MyModel.objects.get(pk=123)
print model_instance.extra_attribute # output: "It's working!"
Django's ForeignKey model field is doing a similar thing, so it is possible :P
I think ForeignKey field is using the contribute_to_class method.
You can create a special Proxy class that will replace the field. When getting or setting field value, you can use 'obj' attribute to access model instance. See this example:
class ObjectField(models.PositiveSmallIntegerField):
class ObjectProxy:
def __init__(self, field):
self.field = field
def __get__(self, obj, model):
if obj is None:
return self.field # this is required for initializing model field
value = obj.__dict__[self.field.name] # get actual field value
# ... here you can do something with value and model instance ("obj")
return value
def __set__(self, obj, value):
# same here
obj.__dict__[self.field.name] = value
def contribute_to_class(self, cls, name):
super().contribute_to_class(cls, name)
# set up our proxy instead of usual field
setattr(cls, name, ObjectField.ObjectProxy(self))
You do not have access to the model instance from inside your Field object, sorry. Django's ForeignKey accomplishes the foo_id thing by having separate name and attname fields, but the actual setting of foo_id = 123 is done the same way as all the other model fields, deep in the QuerySet code, without interacting with the field classes.
And conceptually, what you're trying to do is a bad idea - action-at-a-distance. What if adding a particular field could cause bugs in unrelated model functionality, say, if an attribute another field was expecting got overridden? It would be difficult to debug, to say the least. I don't know what your underlying goal is, but it should probably be done in model code, not a field class.
Here's a ModelField that does what you want:
https://gist.github.com/1987190
That's actually pretty old (like maybe pre-1.0, don't remember now), had to dust it off a bit - I'm not sure if it still works. But it's definitely doable, hopefullly this gives you an idea.
init is called when Django processes the Model Class, not the Model Instance. So, you can add the attribute to the Model Class (e.g. by using 'add_to_class' http://www.alrond.com/en/2008/may/03/monkey-patching-in-django/ ). To add the attribute to the instance you should override the init of the instance (but I think this is not your case).
How about
model_instance = SomeExtraModel.objects.get(pk=1456)
replacing 1456 with something that makes sense
I've only been using Django for a couple of weeks now, so I may be approaching this all kinds of wrong, but:
I have a base ModelForm that I put some boilerplate stuff in to keep things as DRY as possible, and all of my actual ModelForms just subclass that base form. This is working great for error_css_class = 'error' and required_css_class = 'required' but formfield_callback = add_css_classes isn't working like I would expect it to.
forms.py
# snippet I found
def add_css_classes(f, **kwargs):
field = f.formfield(**kwargs)
if field and 'class' not in field.widget.attrs:
field.widget.attrs['class'] = '%s' % field.__class__.__name__.lower()
return field
class BaseForm(forms.ModelForm):
formfield_callback = add_css_classes # not working
error_css_class = 'error'
required_css_class = 'required'
class Meta:
pass
class TimeLogForm(BaseForm):
# I want the next line to be in the parent class
# formfield_callback = add_css_classes
class Meta(BaseForm.Meta):
model = TimeLog
The end goal is to slap some jquery datetime pickers on forms with a class of datefield/timefield/datetimefield. I want all of the date time fields within the app to use the same widget, so I opted to do it this way than explicitly doing it for each field in every model. Adding an extra line to each form class isn't that big of a deal, but it just bugged me that I couldn't figure it out. Digging around in the django source showed this is probably doing something I'm not understanding:
django.forms.models
class ModelFormMetaclass(type):
def __new__(cls, name, bases, attrs):
formfield_callback = attrs.pop('formfield_callback', None)
But I don't know how __init__ and __new__ are all intermangled. In BaseForm I tried overriding __init__ and setting formfield_callback before and after the call to super, but I'm guessing it needs to be somewhere in args or kwargs.
__new__ is called before object construction. Actually this is a factory method that returns the instance of a newly constructed object.
So there there are 3 key lines in ModelFormMetaclass:
formfield_callback = attrs.pop('formfield_callback', None) #1
fields = fields_for_model(opts.model, opts.fields,
opts.exclude, opts.widgets, formfield_callback) #2
new_class.base_fields = fields #3
In the class we attach base_fields to our form.
Now let's look to ModelForm class:
class ModelForm(BaseModelForm):
__metaclass__ = ModelFormMetaclass
This means that ModelFormMetaclass.__new__(...) will be called when we create a ModelForm instance to change the structure of the future instance. And attrs of __new__ (def __new__(cls, name, bases, attrs)) in ModelFormMetaclass is a dict of all attributes of ModelForm class.
So decision is to create new InheritedFormMetaclass for our case (inheriting it from ModelFormMetaclass). Don't forget to call new of the parent in InheritedFormMetaclass. Then create our BaseForm class and say:
__metaclass__ = InheritedFormMetaclass
In __new__(...) implementation of InheritedFormMetaclass we could do all we want.
If my answer is not detailed enough please let me know with help of comments.
You may set widgets class like this:
class TimeLogForm(BaseForm):
# I want the next line to be in the parent class
# formfield_callback = add_css_classes
class Meta(BaseForm.Meta):
model = TimeLog
widgets = {
'some_fields' : SomeWidgets(attrs={'class' : 'myclass'})
}
For what you're trying to accomplish, I think you're better off just looping through the fields on form init. For example,
class BaseForm(forms.ModelForm):
def __init__(self, *args, **kwargs):
super(BaseForm, self).__init__(*args, **kwargs)
for name, field in self.fields.items():
field.widget.attrs['class'] = 'error'
Clearly you'll need a little more logic for your specific case. If you want to use the approach that sergzach suggested (overkill for your particular problem I think), here's some code for you that will call formfield_callback on the base class in the case the subclass doesn't define it.
baseform_formfield_callback(field):
# do some stuff
return field.formfield()
class BaseModelFormMetaclass(forms.models.ModelFormMetaclass):
def __new__(cls, name, bases, attrs):
if not attrs.has_key('formfield_callback'):
attrs['formfield_callback'] = baseform_formfield_callback
new_class = super(BaseModelFormMetaclass, cls).__new__(
cls, name, bases, attrs)
return new_class
class BaseModelForm(forms.ModelForm):
__metaclass__ = OrganizationModelFormMetaclass
# other form stuff
Finally, you might wanna look into crispy forms: https://github.com/maraujop/django-crispy-forms
sergzach is correct that you have to use metaclasses; overriding __init__ is not enough. The reason is that the metaclass for ModelForm (which will be called for all ModelForm subclasses unless you specify another metaclass in a subclass) takes the class definition, and using the values in the class definition creates a class with class attributes. For example, both META.fields and our formfield_callback is used to create form Fields with various option (like which widget).
That means AFAIU formfield_callback is a parameter to the metaclass used when creating your custom model form class, not some value used at runtime when actual form instances are created. That makes placing formfield_callback in __init__ useless.
I solved a similiar problem with a custom metaclass like
from django.forms.models import ModelFormMetaclass
class MyModelFormMetaclass(ModelFormMetaclass):
def __new__(cls,name,bases,attrs):
attrs['formfield_callback']=my_callback_function
return super(MyModelFormMetaclass,cls).__new__(cls,name,bases,attrs)
and in the base class for all my model forms setting the metaclass
class MyBaseModelForm(ModelForm):
__metaclass__=MyModelFormMetaclass
...
which can be used like (at least in Django 1.6)
class MyConcreteModelForm(MyBaseModelForm):
# no need setting formfield_callback here
...
I am trying to add a dynamic Meta attribute to all of my Django models using model inheritance, but I can't get it to work. I have a permission that I want to add to all my models like this:
class ModelA(models.Model):
class Meta:
permisssions =(('view_modela','Can view Model A'),)
class ModelB(models.Model):
class Meta:
permisssions =(('view_modelb','Can view Model B'),)
I tried creating an abstract base class like this:
class CustomModel(models.Model):
def __init__(self, *args, **kwargs):
self._meta.permissions.append(('view_'+self._meta.module_name, u'Can view %s' % self._meta.verbose_name))
super(CustomModel,self).__init__(*args, **kwargs)
class ModelA(CustomModel):
....
class ModelB(CustomModel):
...
but it's not working. Is this the right approach? Because Django uses introspection to construct the Model classes, I'm not sure if adding permissions during the __init__() of the class will even work. With my current implementation every time I access a model instance it appends another tuple of the permissions.
Your instinct is right that this won't work. In Django, permissions are stored in the database, which means that:
they need to be available at the class level when syncdb is run in order to populate the auth_permission table (and your approach requires an instance, which won't be made during syncdb)
even if you did add it to _meta.permissions in __init__, the User object wouldn't pick it up in any permission check calls because those consult the permissions table in the DB (and a cache of that table, at that).
Your goal can't be accomplished using inheritance. What you actually need here is a Python metaclass.
This metaclass re-writes your ModelA and ModelB class definitions dynamically before they are defined, thus it doesn't require a ModelA instance, and is available to syncdb. Since Django's models also use metaclasses to build the Meta object in the first place, the only requirement is that your metaclass must inherit from the same metaclass as Django's models.
Here's some sample code (Python 2):
from django.db.models.base import ModelBase
class CustomModelMetaClass(ModelBase):
def __new__(cls, name, bases, attrs):
klas = super(CustomModelMetaClass, cls).__new__(cls, name, bases, attrs)
klas._meta.permissions.append(
(
'view_{0.module_name}'.format(klas._meta),
u'Can view {0.verbose_name}'.format(klas._meta))
)
return klas
class ModelA(models.Model):
__metaclass__ = CustomModelMetaClass
test = models.CharField(max_length=5)
Python 3:
from django.db.models.base import ModelBase
class CustomModelMetaClass(ModelBase):
def __new__(cls, name, bases, attrs):
klas = super().__new__(cls, name, bases, attrs)
klas._meta.permissions.append(
(
'view_{0.module_name}'.format(klas._meta),
'Can view {0.verbose_name}'.format(klas._meta))
)
return klas
class ModelA(models.Model, metaclass=CustomModelMetaClass):
test = models.CharField(max_length=5)
Note that permissions in this case will be written only on migrate. If you need to change permissions dynamically at run time base on the user, you'll want to provide your own authentication backend.
Try to use a custom manager:
#create a custom manager
class DynTableNameManager(models.Manager):
#overwrite all() (example)
#provide table_name
def all(self, table_name):
from django.db import connection
cursor = connection.cursor()
cursor.execute("""
SELECT id, name
FROM %s
""" % table_name)
result_list = []
for row in cursor.fetchall():
p = self.model(id=row[0], name=row[1])
result_list.append(p)
return result_list
#cerate a dummy table
class DummyTable(models.Model):
name = models.CharField ( max_length = 200 )
objects = DynTableNameManager()
use like this:
f = DummyTable.objects.all('my_table_name')