I want to create an object that contains 2 links to Users. For example:

class GameClaim(models.Model):
    target = models.ForeignKey(User)
    claimer = models.ForeignKey(User)
    isAccepted = models.BooleanField()

but I am getting the following errors when running the server:

• Accessor for field ‘target’ clashes with related field ‘User.gameclaim_set’. Add a related_name argument to the definition for ‘target’.

• Accessor for field ‘claimer’ clashes with related field ‘User.gameclaim_set’. Add a related_name argument to the definition for ‘claimer’.

Can you please explain why I am getting the errors and how to fix them?

You have two foreign keys to User. Django automatically creates a reverse relation from User back to GameClaim, which is usually gameclaim_set. However, because you have two FKs, you would have two gameclaim_set attributes, which is obviously impossible. So you need to tell Django what name to use for the reverse relation.

Use the related_name attribute in the FK definition. e.g.

class GameClaim(models.Model):
    target = models.ForeignKey(User, related_name='gameclaim_targets')
    claimer = models.ForeignKey(User, related_name='gameclaim_users')
    isAccepted = models.BooleanField()

The User model is trying to create two fields with the same name, one for the GameClaims that have that User as the target, and another for the GameClaims that have that User as the claimer. Here’s the docs on related_name, which is Django’s way of letting you set the names of the attributes so the autogenerated ones don’t conflict.

The OP isn’t using a abstract base class… but if you are, you will find that hard coding the related_name in the FK (e.g. …, related_name=”myname”) will result in a number of these conflict errors – one for each inherited class from the base class. The link provided below contains the workaround, which is simple, but definitely not obvious.

From the django docs…

If you are using the related_name attribute on a ForeignKey or ManyToManyField, you must always specify a unique reverse name for the field. This would normally cause a problem in abstract base classes, since the fields on this class are included into each of the child classes, with exactly the same values for the attributes (including related_name) each time.

More info here.

Sometimes you have to use extra formatting in related_name – actually, any time when inheritance is used.

class Value(models.Model):
    value = models.DecimalField(decimal_places=2, max_digits=5)
    animal = models.ForeignKey(
        Animal, related_name="%(app_label)s_%(class)s_related")

    class Meta:
        abstract = True

class Height(Value):
    pass

class Weigth(Value):
    pass

class Length(Value):
    pass

No clash here, but related_name is defined once and Django will take care for creating unique relation names.

then in children of Value class, you’ll have access to:

herdboard_height_related
herdboard_lenght_related
herdboard_weight_related

I seem to come across this occasionally when I add a submodule as an application to a django project, for example given the following structure:

myapp/
myapp/module/
myapp/module/models.py

If I add the following to INSTALLED_APPS:

'myapp',
'myapp.module',

Django seems to process the myapp.mymodule models.py file twice and throws the above error. This can be resolved by not including the main module in the INSTALLED_APPS list:

'myapp.module',

Including the myapp instead of myapp.module causes all the database tables to be created with incorrect names, so this seems to be the correct way to do it.

I came across this post while looking for a solution to this problem so figured I’d put this here :)

Just adding to Jordan’s answer (thanks for the tip Jordan) it can also happen if you import the level above the apps and then import the apps e.g.

myproject/ apps/ foo_app/ bar_app/

So if you are importing apps, foo_app and bar_app then you could get this issue. I had apps, foo_app and bar_app all listed in settings.INSTALLED_APPS

And you want to avoid importing apps anyway, because then you have the same app installed in 2 different namespaces

apps.foo_app and foo_app

I’m trying to set up my uploads so that if user joe uploads a file it goes to MEDIA_ROOT/joe as opposed to having everyone’s files go to MEDIA_ROOT. The problem is I don’t know how to define this in the model. Here is how it currently looks:

class Content(models.Model):
    name = models.CharField(max_length=200)
    user = models.ForeignKey(User)
    file = models.FileField(upload_to='.')

So what I want is instead of ‘.’ as the upload_to, have it be the user’s name.

I understand that as of Django 1.0 you can define your own function to handle the upload_to but that function has no idea of who the user will be either so I’m a bit lost.

Thanks for the help!

You’ve probably read the documentation, so here’s an easy example to make it make sense:

def content_file_name(instance, filename):
    return '/'.join(['content', instance.user.username, filename])

class Content(models.Model):
    name = models.CharField(max_length=200)
    user = models.ForeignKey(User)
    file = models.FileField(upload_to=content_file_name)

As you can see, you don’t even need to use the filename given – you could override that in your upload_to callable too if you liked.

This really helped. For a bit more brevity’s sake, decided to use lambda in my case:

file = models.FileField(
    upload_to=lambda instance, filename: '/'.join(['mymodel', str(instance.pk), filename]),
)

A note on using the ‘instance’ object’s pk value. According to the documentation:

In most cases, this object will not have been saved to the database yet, so if it uses the default AutoField, it might not yet have a value for its primary key field.

Therefore the validity of using pk depends on how your particular model is defined.

If you have problems with migrations you probably should be using @deconstructible decorator.

import datetime
import os
import unicodedata

from django.core.files.storage import default_storage
from django.utils.deconstruct import deconstructible
from django.utils.encoding import force_text, force_str


@deconstructible
class UploadToPath(object):
    def __init__(self, upload_to):
        self.upload_to = upload_to

    def __call__(self, instance, filename):
        return self.generate_filename(filename)

    def get_directory_name(self):
        return os.path.normpath(force_text(datetime.datetime.now().strftime(force_str(self.upload_to))))

    def get_filename(self, filename):
        filename = default_storage.get_valid_name(os.path.basename(filename))
        filename = force_text(filename)
        filename = unicodedata.normalize('NFKD', filename).encode('ascii', 'ignore').decode('ascii')
        return os.path.normpath(filename)

    def generate_filename(self, filename):
        return os.path.join(self.get_directory_name(), self.get_filename(filename))

Usage:

class MyModel(models.Model):
    file = models.FileField(upload_to=UploadToPath('files/%Y/%m/%d'), max_length=255)

I want to create a model object, like Person, if person’s id doesn’t not exist, or I will get that person object.

The code to create a new person as following:

class Person(models.Model):
    identifier = models.CharField(max_length = 10)
    name = models.CharField(max_length = 20)
    objects = PersonManager()

class PersonManager(models.Manager):
    def create_person(self, identifier):
        person = self.create(identifier = identifier)
        return person

But I don’t know where to check and get the existing person object.

If you’re looking for “update if exists else create” use case, please refer to @Zags excellent answer

Django already has a get_or_create, https://docs.djangoproject.com/en/dev/ref/models/querysets/#get-or-create

For you it could be :

id = 'some identifier'
person, created = Person.objects.get_or_create(identifier=id)

if created:
   # means you have created a new person
else:
   # person just refers to the existing one

It’s unclear whether your question is asking for the get_or_create method (available from at least Django 1.3) or the update_or_create method (new in Django 1.7). It depends on how you want to update the user object.

Sample use is as follows:

# In both cases, the call will get a person object with matching
# identifier or create one if none exists; if a person is created,
# it will be created with name equal to the value in `name`.

# In this case, if the Person already exists, its existing name is preserved
person, created = Person.objects.get_or_create(
        identifier=identifier, defaults={"name": name}
)

# In this case, if the Person already exists, its name is updated
person, created = Person.objects.update_or_create(
        identifier=identifier, defaults={"name": name}
)

Django has support for this, check get_or_create

person, created = Person.objects.get_or_create(name='abc')
if created:
    # A new person object created
else:
    # person object already exists

For only a small amount of objects the update_or_create works well, but if you’re doing over a large collection it won’t scale well. update_or_create always first runs a SELECT and thereafter an UPDATE.

for the_bar in bars:
    updated_rows = SomeModel.objects.filter(bar=the_bar).update(foo=100)
        if not updated_rows:
            # if not exists, create new
            SomeModel.objects.create(bar=the_bar, foo=100)

This will at best only run the first update-query, and only if it matched zero rows run another INSERT-query. Which will greatly increase your performance if you expect most of the rows to actually be existing.

It all comes down to your use case though. If you are expecting mostly inserts then perhaps the bulk_create() command could be an option.

Thought I’d add an answer since your question title looks like it is asking how to create or update, rather than get or create as described in the question body.

If you did want to create or update an object, the .save() method already has this behaviour by default, from the docs:

Django abstracts the need to use INSERT or UPDATE SQL statements. Specifically, when you call save(), Django follows this algorithm:

If the object’s primary key attribute is set to a value that evaluates to True (i.e., a value other than None or the empty string), Django executes an UPDATE. If the object’s primary key attribute is not set or if the UPDATE didn’t update anything, Django executes an INSERT.

It’s worth noting that when they say ‘if the UPDATE didn’t update anything’ they are essentially referring to the case where the id you gave the object doesn’t already exist in the database.

If one of the input when you create is a primary key, this will be enough:

Person.objects.get_or_create(id=1)

It will automatically update if exist since two data with the same primary key is not allowed.

what I’m trying to do is this:

• get the 30 Authors with highest score ( Author.objects.order_by('-score')[:30] )

• order the authors by last_name

Any suggestions?

What about

import operator

auths = Author.objects.order_by('-score')[:30]
ordered = sorted(auths, key=operator.attrgetter('last_name'))

In Django 1.4 and newer you can order by providing multiple fields.
Reference: https://docs.djangoproject.com/en/dev/ref/models/querysets/#order-by

order_by(*fields)

By default, results returned by a QuerySet are ordered by the ordering tuple given by the ordering option in the model’s Meta. You can override this on a per-QuerySet basis by using the order_by method.

Example:

ordered_authors = Author.objects.order_by('-score', 'last_name')[:30]

The result above will be ordered by score descending, then by last_name ascending. The negative sign in front of "-score" indicates descending order. Ascending order is implied.

I just wanted to illustrate that the built-in solutions (SQL-only) are not always the best ones. At first I thought that because Django’s QuerySet.objects.order_by method accepts multiple arguments, you could easily chain them:

ordered_authors = Author.objects.order_by('-score', 'last_name')[:30]

But, it does not work as you would expect. Case in point, first is a list of presidents sorted by score (selecting top 5 for easier reading):

>>> auths = Author.objects.order_by('-score')[:5]
>>> for x in auths: print x
... 
James Monroe (487)
Ulysses Simpson (474)
Harry Truman (471)
Benjamin Harrison (467)
Gerald Rudolph (464)

Using Alex Martelli’s solution which accurately provides the top 5 people sorted by last_name:

>>> for x in sorted(auths, key=operator.attrgetter('last_name')): print x
... 
Benjamin Harrison (467)
James Monroe (487)
Gerald Rudolph (464)
Ulysses Simpson (474)
Harry Truman (471)

And now the combined order_by call:

>>> myauths = Author.objects.order_by('-score', 'last_name')[:5]
>>> for x in myauths: print x
... 
James Monroe (487)
Ulysses Simpson (474)
Harry Truman (471)
Benjamin Harrison (467)
Gerald Rudolph (464)

As you can see it is the same result as the first one, meaning it doesn’t work as you would expect.

Here’s a way that allows for ties for the cut-off score.

author_count = Author.objects.count()
cut_off_score = Author.objects.order_by('-score').values_list('score')[min(30, author_count)]
top_authors = Author.objects.filter(score__gte=cut_off_score).order_by('last_name')

You may get more than 30 authors in top_authors this way and the min(30,author_count) is there incase you have fewer than 30 authors.

I’m wondering if it’s possible to define a foreign key in a models.py file in Django that is a reference to a table in another app?

In other words, I have two apps, called cf and profiles, and in cf/models.py I have (amongst other things):

class Movie(models.Model):
    title = models.CharField(max_length=255)

and in profiles/models.py I want to have:

class MovieProperty(models.Model):
    movie = models.ForeignKey(Movie)

But I can’t get it to work. I’ve tried:

    movie = models.ForeignKey(cf.Movie)

and I’ve tried importing cf.Movie at the beginning of models.py, but I always get errors, such as:

NameError: name 'User' is not defined

Am I breaking the rules by trying to tie two apps together in this way, or have I just got the syntax wrong?

According to the docs, your second attempt should work:

To refer to models defined in another application, you must instead explicitly specify the application label. For example, if the Manufacturer model above is defined in another application called production, you’d need to use:

class Car(models.Model):
    manufacturer = models.ForeignKey('production.Manufacturer')

Have you tried putting it into quotes?

It is also possible to pass the class itself:

from django.db import models
from production import models as production_models

class Car(models.Model):
    manufacturer = models.ForeignKey(production_models.Manufacturer)

OK – I’ve figured it out. You can do it, you just have to use the right import syntax. The correct syntax is:

from prototype.cf.models import Movie

My mistake was not specifying the .models part of that line. D’oh!

Here’s my model. What I want to do is generate a new file and overwrite the existing one whenever a model instance is saved:

class Kitten(models.Model):
    claw_size = ...
    license_file = models.FileField(blank=True, upload_to='license')

    def save(self, *args, **kwargs):
        #Generate a new license file overwriting any previous version
        #and update file path
        self.license_file = ???
        super(Request,self).save(*args, **kwargs)

I see lots of documentation about how to upload a file. But how do I generate a file, assign it to a model field and have Django store it in the right place?

You want to have a look at FileField and FieldFile in the Django docs, and especially FieldFile.save().

Basically, a field declared as a FileField, when accessed, gives you an instance of class FieldFile, which gives you several methods to interact with the underlying file. So, what you need to do is:

self.license_file.save(new_name, new_contents)

where new_name is the filename you wish assigned and new_contents is the content of the file. Note that new_contents must be an instance of either django.core.files.File or django.core.files.base.ContentFile (see given links to manual for the details). The two choices boil down to:

# Using File
f = open('/path/to/file')
self.license_file.save(new_name, File(f))
# Using ContentFile
self.license_file.save(new_name, ContentFile('A string with the file content'))

Accepted answer is certainly a good solution, but here is the way I went about generating a CSV and serving it from a view.

Thought it was worth while putting this here as it took me a little bit of fiddling to get all the desirable behaviour (overwrite existing file, storing to the right spot, not creating duplicate files etc).

Django 1.4.1

Python 2.7.3

#Model
class MonthEnd(models.Model):
    report = models.FileField(db_index=True, upload_to='not_used')

import csv
from os.path import join

#build and store the file
def write_csv():
    path = join(settings.MEDIA_ROOT, 'files', 'month_end', 'report.csv')
    f = open(path, "w+b")

    #wipe the existing content
    f.truncate()

    csv_writer = csv.writer(f)
    csv_writer.writerow(('col1'))

    for num in range(3):
        csv_writer.writerow((num, ))

    month_end_file = MonthEnd()
    month_end_file.report.name = path
    month_end_file.save()

from my_app.models import MonthEnd

#serve it up as a download
def get_report(request):
    month_end = MonthEnd.objects.get(file_criteria=criteria)

    response = HttpResponse(month_end.report, content_type='text/plain')
    response['Content-Disposition'] = 'attachment; filename=report.csv'

    return response

It’s good practice to use a context manager or call close() in case of exceptions during the file saving process. Could happen if your storage backend is down, etc.

Any overwrite behavior should be configured in your storage backend. For example S3Boto3Storage has a setting AWS_S3_FILE_OVERWRITE. If you’re using FileSystemStorage you can write a custom mixin.

You might also want to call the model’s save method instead of the FileField’s save method if you want any custom side-effects to happen, like last-updated timestamps. If that’s the case, you can also set the name attribute of the file to the name of the file – which is relative to MEDIA_ROOT. It defaults to the full path of the file which can cause problems if you don’t set it – see File.__init__() and File.name.

Here’s an example where self is the model instance where my_file is the FileField / ImageFile, calling save() on the whole model instance instead of just FileField:

import os
from django.core.files import File

with open(filepath, 'rb') as fi:
    self.my_file = File(fi, name=os.path.basename(fi.name))
    self.save()

When you have a model field with a choices option you tend to have some magic values associated with human readable names. Is there in Django a convenient way to set these fields by the human readable name instead of the value?

Consider this model:

class Thing(models.Model):
  PRIORITIES = (
    (0, 'Low'),
    (1, 'Normal'),
    (2, 'High'),
  )

  priority = models.IntegerField(default=0, choices=PRIORITIES)

At some point we have a Thing instance and we want to set its priority. Obviously you could do,

thing.priority = 1

But that forces you to memorize the Value-Name mapping of PRIORITIES. This doesn’t work:

thing.priority = 'Normal' # Throws ValueError on .save()

Currently I have this silly workaround:

thing.priority = dict((key,value) for (value,key) in Thing.PRIORITIES)['Normal']

but that’s clunky. Given how common this scenario could be I was wondering if anyone had a better solution. Is there some field method for setting fields by choice name which I totally overlooked?

Do as seen here. Then you can use a word that represents the proper integer.

Like so:

LOW = 0
NORMAL = 1
HIGH = 2
STATUS_CHOICES = (
    (LOW, 'Low'),
    (NORMAL, 'Normal'),
    (HIGH, 'High'),
)

Then they are still integers in the DB.

Usage would be thing.priority = Thing.NORMAL

I’d probably set up the reverse-lookup dict once and for all, but if I hadn’t I’d just use:

thing.priority = next(value for value, name in Thing.PRIORITIES
                      if name=='Normal')

which seems simpler than building the dict on the fly just to toss it away again;-).

Here’s a field type I wrote a few minutes ago that I think does what you want. Its constructor requires an argument ‘choices’, which may be either a tuple of 2-tuples in the same format as the choices option to IntegerField, or instead a simple list of names (ie ChoiceField((‘Low’, ‘Normal’, ‘High’), default=’Low’) ). The class takes care of the mapping from string to int for you, you never see the int.

  class ChoiceField(models.IntegerField):
    def __init__(self, choices, **kwargs):
        if not hasattr(choices[0],'__iter__'):
            choices = zip(range(len(choices)), choices)

        self.val2choice = dict(choices)
        self.choice2val = dict((v,k) for k,v in choices)

        kwargs['choices'] = choices
        super(models.IntegerField, self).__init__(**kwargs)

    def to_python(self, value):
        return self.val2choice[value]

    def get_db_prep_value(self, choice):
        return self.choice2val[choice]

I appreciate the constant defining way but I believe Enum type is far best for this task. They can represent integer and a string for an item in the same time, while keeping your code more readable.

Enums were introduced to Python in version 3.4. If you are using any lower (such as v2.x) you can still have it by installing the backported package: pip install enum34.

# myapp/fields.py
from enum import Enum    


class ChoiceEnum(Enum):

    @classmethod
    def choices(cls):
        choices = list()

        # Loop thru defined enums
        for item in cls:
            choices.append((item.value, item.name))

        # return as tuple
        return tuple(choices)

    def __str__(self):
        return self.name

    def __int__(self):
        return self.value


class Language(ChoiceEnum):
    Python = 1
    Ruby = 2
    Java = 3
    PHP = 4
    Cpp = 5

# Uh oh
Language.Cpp._name_ = 'C++'

This is pretty much all. You can inherit the ChoiceEnum to create your own definitions and use them in a model definition like:

from django.db import models
from myapp.fields import Language

class MyModel(models.Model):
    language = models.IntegerField(choices=Language.choices(), default=int(Language.Python))
    # ...

Querying is icing on the cake as you may guess:

MyModel.objects.filter(language=int(Language.Ruby))
# or if you don't prefer `__int__` method..
MyModel.objects.filter(language=Language.Ruby.value)

Representing them in string is also made easy:

# Get the enum item
lang = Language(some_instance.language)

print(str(lang))
# or if you don't prefer `__str__` method..
print(lang.name)

# Same as get_FOO_display
lang.name == some_instance.get_language_display()

class Sequence(object):
    def __init__(self, func, *opts):
        keys = func(len(opts))
        self.attrs = dict(zip([t[0] for t in opts], keys))
        self.choices = zip(keys, [t[1] for t in opts])
        self.labels = dict(self.choices)
    def __getattr__(self, a):
        return self.attrs[a]
    def __getitem__(self, k):
        return self.labels[k]
    def __len__(self):
        return len(self.choices)
    def __iter__(self):
        return iter(self.choices)
    def __deepcopy__(self, memo):
        return self

class Enum(Sequence):
    def __init__(self, *opts):
        return super(Enum, self).__init__(range, *opts)

class Flags(Sequence):
    def __init__(self, *opts):
        return super(Flags, self).__init__(lambda l: [1<<i for i in xrange(l)], *opts)

Use it like this:

Priorities = Enum(
    ('LOW', 'Low'),
    ('NORMAL', 'Normal'),
    ('HIGH', 'High')
)

priority = models.IntegerField(default=Priorities.LOW, choices=Priorities)

As of Django 3.0, you can use:

class ThingPriority(models.IntegerChoices):
    LOW = 0, 'Low'
    NORMAL = 1, 'Normal'
    HIGH = 2, 'High'


class Thing(models.Model):
    priority = models.IntegerField(default=ThingPriority.LOW, choices=ThingPriority.choices)

# then in your code
thing = get_my_thing()
thing.priority = ThingPriority.HIGH

Simply replace your numbers with the human readable values you would like. As such:

PRIORITIES = (
('LOW', 'Low'),
('NORMAL', 'Normal'),
('HIGH', 'High'),
)

This makes it human readable, however, you’d have to define your own ordering.

My answer is very late and might seem obvious to nowadays-Django experts, but to whoever lands here, i recently discovered a very elegant solution brought by django-model-utils: https://django-model-utils.readthedocs.io/en/latest/utilities.html#choices

This package allows you to define Choices with three-tuples where:

• The first item is the database value
• The second item is a code-readable value
• The third item is a human-readable value

So here’s what you can do:

from model_utils import Choices

class Thing(models.Model):
    PRIORITIES = Choices(
        (0, 'low', 'Low'),
        (1, 'normal', 'Normal'),
        (2, 'high', 'High'),
      )

    priority = models.IntegerField(default=PRIORITIES.normal, choices=PRIORITIES)

thing.priority = getattr(Thing.PRIORITIES.Normal)

This way:

• You can use your human-readable value to actually choose the value of your field (in my case, it’s useful because i’m scraping wild content and storing it in a normalized way)
• A clean value is stored in your database
• You have nothing non-DRY to do ;)

Enjoy :)

Originally I used a modified version of @Allan’s answer:

from enum import IntEnum, EnumMeta

class IntegerChoiceField(models.IntegerField):
    def __init__(self, choices, **kwargs):
        if hasattr(choices, '__iter__') and isinstance(choices, EnumMeta):
            choices = list(zip(range(1, len(choices) + 1), [member.name for member in list(choices)]))

        kwargs['choices'] = choices
        super(models.IntegerField, self).__init__(**kwargs)

    def to_python(self, value):
        return self.choices(value)

    def get_db_prep_value(self, choice):
        return self.choices[choice]

models.IntegerChoiceField = IntegerChoiceField

GEAR = IntEnum('GEAR', 'HEAD BODY FEET HANDS SHIELD NECK UNKNOWN')

class Gear(Item, models.Model):
    # Safe to assume last element is largest value member of an enum?
    #type = models.IntegerChoiceField(GEAR, default=list(GEAR)[-1].name)
    largest_member = GEAR(max([member.value for member in list(GEAR)]))
    type = models.IntegerChoiceField(GEAR, default=largest_member)

    def __init__(self, *args, **kwargs):
        super(Gear, self).__init__(*args, **kwargs)

        for member in GEAR:
            setattr(self, member.name, member.value)

print(Gear().HEAD, (Gear().HEAD == GEAR.HEAD.value))

Simplified with the django-enumfields package package which I now use:

from enumfields import EnumIntegerField, IntEnum

GEAR = IntEnum('GEAR', 'HEAD BODY FEET HANDS SHIELD NECK UNKNOWN')

class Gear(Item, models.Model):
    # Safe to assume last element is largest value member of an enum?
    type = EnumIntegerField(GEAR, default=list(GEAR)[-1])
    #largest_member = GEAR(max([member.value for member in list(GEAR)]))
    #type = EnumIntegerField(GEAR, default=largest_member)

    def __init__(self, *args, **kwargs):
        super(Gear, self).__init__(*args, **kwargs)

        for member in GEAR:
            setattr(self, member.name, member.value)

Model’s choices option accepts a sequence consisting itself of iterables of exactly two items (e.g. [(A, B), (A, B) …]) to use as choices for this field.

In addition, Django provides enumeration types that you can subclass to define choices in a concise way:

class ThingPriority(models.IntegerChoices):
    LOW = 0, _('Low')
    NORMAL = 1, _('Normal')
    HIGH = 2, _('High')

class Thing(models.Model):
    priority = models.IntegerField(default=ThingPriority.NORMAL, choices=ThingPriority.choices)

Django supports adding an extra string value to the end of this tuple to be used as the human-readable name, or label. The label can be a lazy translatable string.

   # in your code 
   thing = get_thing() # instance of Thing
   thing.priority = ThingPriority.LOW

Note: you can use that using ThingPriority.HIGH, ThingPriority.['HIGH'], or ThingPriority(0) to access or lookup enum members.