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Django: Get list of model fields?

I've defined a User class which (ultimately) inherits from models.Model. I want to get a list of all the fields defined for this model. For example, phone_number = CharField(max_length=20). Basically, I want to retrieve anything that inherits from the Field class.

I thought I'd be able to retrieve these by taking advantage of inspect.getmembers(model), but the list it returns doesn't contain any of these fields. It looks like Django has already gotten a hold of the class and added all its magic attributes and stripped out what's actually been defined. So... how can I get these fields? They probably have a function for retrieving them for their own internal purposes?


d
daaawx

Django versions 1.8 and later:

You should use get_fields():

[f.name for f in MyModel._meta.get_fields()]

The get_all_field_names() method is deprecated starting from Django 1.8 and will be removed in 1.10.

The documentation page linked above provides a fully backwards-compatible implementation of get_all_field_names(), but for most purposes the previous example should work just fine.

Django versions before 1.8:

model._meta.get_all_field_names()

That should do the trick.

That requires an actual model instance. If all you have is a subclass of django.db.models.Model, then you should call myproject.myapp.models.MyModel._meta.get_all_field_names()


Wanted the objects too, not just their names. This seems to be available in model._meta.fields though, and their names are retrievable with field.name it seems. I just hope this is the most stable way to retrieve this info :)
not entirely sure. The underscore seems to indicate it's an internal API, would be cool if the Django guys promoted this up to an actually public method call on django.db.models.Model. I'll dig into it and see what I can find
I guess doing it via the _meta attribute is the only way... Additionally look into _meta.many_to_many for ManyToMany fields!
It's a good solution but this method includes the reverse relation fields such as a reverse ForeignKey and those are not exactly "fields". Anyone know how to distinguish the actual Fields?
@rossipedia: Just noting that the ._meta API is public (though it used to be private, probably when this answer was first written): docs.djangoproject.com/en/1.8/ref/models/meta/…
b
bignose

As most of answers are outdated I'll try to update you on Django 2.2 Here posts- your app (posts, blog, shop, etc.)

1) From model link: https://docs.djangoproject.com/en/stable/ref/models/meta/

from posts.model import BlogPost

all_fields = BlogPost._meta.fields
#or
all_fields = BlogPost._meta.get_fields()

Note that:

all_fields=BlogPost._meta.get_fields()

Will also get some relationships, which, for ex: you can not display in a view. As in my case:

Organisation._meta.fields
(<django.db.models.fields.AutoField: id>, <django.db.models.fields.DateField: created>...

and

Organisation._meta.get_fields()
(<ManyToOneRel: crm.activity>, <django.db.models.fields.AutoField: id>, <django.db.models.fields.DateField: created>...

2) From instance

from posts.model import BlogPost

bp = BlogPost()
all_fields = bp._meta.fields

3) From parent model

Let's suppose that we have Post as the parent model and you want to see all the fields in a list, and have the parent fields to be read-only in Edit mode.

from django.contrib import admin
from posts.model import BlogPost 

@admin.register(BlogPost)
class BlogPost(admin.ModelAdmin):
    all_fields = [f.name for f in Organisation._meta.fields]
    parent_fields = BlogPost.get_deferred_fields(BlogPost)

    list_display = all_fields
    read_only = parent_fields

I'll assume this is correct and accept it as the new answer. Thanks for the update!
@mpen Wont claim it the best way or the most pythonic but below will/should get the values u would want to display in a view, so the headers of an HTML table if u will. As get_fields() returns a tuple, u can iterate over it and get the values that look like appname.Model.field_name, below cleans up the values from the second dot, includes the case in which an underscore was used in the field name as well as make them a title, so modify as needed for each unique situation. clean_field_names = [str(h).split('.')[2].replace("_", " ").title() for h in all_fields]
The from instance example should be: all_fields = bp._meta.fields
@GeorgeKettleborough i see the point. Doesn't it work from class directly also? Don't have the example to test on. Anyway the example is about instance, so it should be from bp or at least bp.__class__
W
Wil

The get_all_related_fields() method mentioned herein has been deprecated in 1.8. From now on it's get_fields().

>> from django.contrib.auth.models import User
>> User._meta.get_fields()

This should now be the accepted answer. Terse, and to the point.
Yes this is the answer.
V
Visgean Skeloru

I find adding this to django models quite helpful:

def __iter__(self):
    for field_name in self._meta.get_all_field_names():
        value = getattr(self, field_name, None)
        yield (field_name, value)

This lets you do:

for field, val in object:
    print field, val

What about ForeignKey? I have errors like this django.db.models.fields.related.RelatedObjectDoesNotExist: CustomModel has no custom_attribute.
ForeignKey work fine for me. Although, silently catching all exceptions is an anti-pattern. Much better to catch AttributeError, or at least log that some exception was silently swallowed.
self._meta.get_all_field_names() has been depreciated and removed. You can use something like for field in self._meta.get_fields() and then yield (field.name, field.value_from_object(self))
R
Rockallite

This does the trick. I only test it in Django 1.7.

your_fields = YourModel._meta.local_fields
your_field_names = [f.name for f in your_fields]

Model._meta.local_fields does not contain many-to-many fields. You should get them using Model._meta.local_many_to_many.


N
Nader Alexan

It is not clear whether you have an instance of the class or the class itself and trying to retrieve the fields, but either way, consider the following code

Using an instance

instance = User.objects.get(username="foo")
instance.__dict__ # returns a dictionary with all fields and their values
instance.__dict__.keys() # returns a dictionary with all fields
list(instance.__dict__.keys()) # returns list with all fields

Using a class

User._meta.__dict__.get("fields") # returns the fields

# to get the field names consider looping over the fields and calling __str__()
for field in User._meta.__dict__.get("fields"):
    field.__str__() # e.g. 'auth.User.id'

V
Vivek Anand
def __iter__(self):
    field_names = [f.name for f in self._meta.fields]
    for field_name in field_names:
        value = getattr(self, field_name, None)
        yield (field_name, value)

This worked for me in django==1.11.8


a
aboutaaron

MyModel._meta.get_all_field_names() was deprecated several versions back and removed in Django 1.10.

Here's the backwards-compatible suggestion from the docs:

from itertools import chain

list(set(chain.from_iterable(
    (field.name, field.attname) if hasattr(field, 'attname') else (field.name,)
    for field in MyModel._meta.get_fields()
    # For complete backwards compatibility, you may want to exclude
    # GenericForeignKey from the results.
    if not (field.many_to_one and field.related_model is None)
)))

a
arrt_

Just to add, I am using self object, this worked for me:

[f.name for f in self.model._meta.get_fields()]

k
keepAlive

At least with Django 1.9.9 -- the version I'm currently using --, note that .get_fields() actually also "considers" any foreign model as a field, which may be problematic. Say you have:

class Parent(models.Model):
    id = UUIDField(primary_key=True)

class Child(models.Model):
    parent = models.ForeignKey(Parent)

It follows that

>>> map(lambda field:field.name, Parent._model._meta.get_fields())
['id', 'child']

while, as shown by @Rockallite

>>> map(lambda field:field.name, Parent._model._meta.local_fields)
['id']

C
Carl Brubaker

So before I found this post, I successfully found this to work.

Model._meta.fields

It works equally as

Model._meta.get_fields()

I'm not sure what the difference is in the results, if there is one. I ran this loop and got the same output.

for field in Model._meta.fields:
    print(field.name)

A
Anselmo Blanco Dominguez

A detail not mentioned by others:

[f.name for f in MyModel._meta.get_fields()]

get, for example

['id', 'name', 'occupation']

and

[f.get_attname() for f in MyModel._meta.get_fields()]

get

['id', 'name', 'occupation_id']

If

reg = MyModel.objects.first()

then

reg.occupation

get, for example

<Occupation: Dev>

and

reg.occupation_id

get

1

R
Ramesh Ponnusamy

In sometimes we need the db columns as well:

def get_db_field_names(instance):
   your_fields = instance._meta.local_fields
   db_field_names=[f.name+'_id' if f.related_model is not None else f.name  for f in your_fields]
   model_field_names = [f.name for f in your_fields]
   return db_field_names,model_field_names

Call the method to get the fields:

db_field_names,model_field_names=get_db_field_names(Mymodel)

T
Tobias Ernst

Combined multiple answers of the given thread (thanks!) and came up with the following generic solution:

class ReadOnlyBaseModelAdmin(ModelAdmin):
    def has_add_permission(self, request):
        return request.user.is_superuser

    def has_delete_permission(self, request, obj=None):
        return request.user.is_superuser

    def get_readonly_fields(self, request, obj=None):
        return [f.name for f in self.model._meta.get_fields()]

A
Adam Pawluczuk

Why not just use that:

manage.py inspectdb

Example output:

class GuardianUserobjectpermission(models.Model):
    id = models.IntegerField(primary_key=True)  # AutoField?
    object_pk = models.CharField(max_length=255)
    content_type = models.ForeignKey(DjangoContentType, models.DO_NOTHING)
    permission = models.ForeignKey(AuthPermission, models.DO_NOTHING)
    user = models.ForeignKey(CustomUsers, models.DO_NOTHING)

    class Meta:
        managed = False
        db_table = 'guardian_userobjectpermission'
        unique_together = (('user', 'permission', 'object_pk'),)

This isn't an answer to the question.