0
0
mirror of https://github.com/django/django.git synced 2024-11-22 03:18:31 +01:00
django/docs/topics/files.txt

275 lines
8.6 KiB
Plaintext

==============
Managing files
==============
This document describes Django's file access APIs for files such as those
uploaded by a user. The lower level APIs are general enough that you could use
them for other purposes. If you want to handle "static files" (JS, CSS, etc.),
see :doc:`/howto/static-files/index`.
By default, Django stores files locally, using the :setting:`MEDIA_ROOT` and
:setting:`MEDIA_URL` settings. The examples below assume that you're using these
defaults.
However, Django provides ways to write custom `file storage systems`_ that
allow you to completely customize where and how Django stores files. The
second half of this document describes how these storage systems work.
.. _file storage systems: `File storage`_
Using files in models
=====================
When you use a :class:`~django.db.models.FileField` or
:class:`~django.db.models.ImageField`, Django provides a set of APIs you can use
to deal with that file.
Consider the following model, using an :class:`~django.db.models.ImageField` to
store a photo::
from django.db import models
class Car(models.Model):
name = models.CharField(max_length=255)
price = models.DecimalField(max_digits=5, decimal_places=2)
photo = models.ImageField(upload_to="cars")
specs = models.FileField(upload_to="specs")
Any ``Car`` instance will have a ``photo`` attribute that you can use to get at
the details of the attached photo:
.. code-block:: pycon
>>> car = Car.objects.get(name="57 Chevy")
>>> car.photo
<ImageFieldFile: cars/chevy.jpg>
>>> car.photo.name
'cars/chevy.jpg'
>>> car.photo.path
'/media/cars/chevy.jpg'
>>> car.photo.url
'https://media.example.com/cars/chevy.jpg'
This object -- ``car.photo`` in the example -- is a ``File`` object, which means
it has all the methods and attributes described below.
.. note::
The file is saved as part of saving the model in the database, so the actual
file name used on disk cannot be relied on until after the model has been
saved.
For example, you can change the file name by setting the file's
:attr:`~django.core.files.File.name` to a path relative to the file storage's
location (:setting:`MEDIA_ROOT` if you are using the default
:class:`~django.core.files.storage.FileSystemStorage`):
.. code-block:: pycon
>>> import os
>>> from django.conf import settings
>>> initial_path = car.photo.path
>>> car.photo.name = "cars/chevy_ii.jpg"
>>> new_path = settings.MEDIA_ROOT + car.photo.name
>>> # Move the file on the filesystem
>>> os.rename(initial_path, new_path)
>>> car.save()
>>> car.photo.path
'/media/cars/chevy_ii.jpg'
>>> car.photo.path == new_path
True
To save an existing file on disk to a :class:`~django.db.models.FileField`:
.. code-block:: pycon
>>> from pathlib import Path
>>> from django.core.files import File
>>> path = Path("/some/external/specs.pdf")
>>> car = Car.objects.get(name="57 Chevy")
>>> with path.open(mode="rb") as f:
... car.specs = File(f, name=path.name)
... car.save()
...
.. note::
While :class:`~django.db.models.ImageField` non-image data attributes, such
as ``height``, ``width``, and ``size`` are available on the instance, the
underlying image data cannot be used without reopening the image. For
example:
.. code-block:: pycon
>>> from PIL import Image
>>> car = Car.objects.get(name="57 Chevy")
>>> car.photo.width
191
>>> car.photo.height
287
>>> image = Image.open(car.photo)
# Raises ValueError: seek of closed file.
>>> car.photo.open()
<ImageFieldFile: cars/chevy.jpg>
>>> image = Image.open(car.photo)
>>> image
<PIL.JpegImagePlugin.JpegImageFile image mode=RGB size=191x287 at 0x7F99A94E9048>
The ``File`` object
===================
Internally, Django uses a :class:`django.core.files.File` instance any time it
needs to represent a file.
Most of the time you'll use a ``File`` that Django's given you (i.e. a file
attached to a model as above, or perhaps an uploaded file).
If you need to construct a ``File`` yourself, the easiest way is to create one
using a Python built-in ``file`` object:
.. code-block:: pycon
>>> from django.core.files import File
# Create a Python file object using open()
>>> f = open("/path/to/hello.world", "w")
>>> myfile = File(f)
Now you can use any of the documented attributes and methods
of the :class:`~django.core.files.File` class.
Be aware that files created in this way are not automatically closed.
The following approach may be used to close files automatically:
.. code-block:: pycon
>>> from django.core.files import File
# Create a Python file object using open() and the with statement
>>> with open("/path/to/hello.world", "w") as f:
... myfile = File(f)
... myfile.write("Hello World")
...
>>> myfile.closed
True
>>> f.closed
True
Closing files is especially important when accessing file fields in a loop
over a large number of objects. If files are not manually closed after
accessing them, the risk of running out of file descriptors may arise. This
may lead to the following error:
.. code-block:: pytb
OSError: [Errno 24] Too many open files
File storage
============
Behind the scenes, Django delegates decisions about how and where to store files
to a file storage system. This is the object that actually understands things
like file systems, opening and reading files, etc.
Django's default file storage is
``'``:class:`django.core.files.storage.FileSystemStorage`\ ``'``. If you don't
explicitly provide a storage system in the ``default`` key of the
:setting:`STORAGES` setting, this is the one that will be used.
See below for details of the built-in default file storage system, and see
:doc:`/howto/custom-file-storage` for information on writing your own file
storage system.
Storage objects
---------------
Though most of the time you'll want to use a ``File`` object (which delegates to
the proper storage for that file), you can use file storage systems directly.
You can create an instance of some custom file storage class, or -- often more
useful -- you can use the global default storage system:
.. code-block:: pycon
>>> from django.core.files.base import ContentFile
>>> from django.core.files.storage import default_storage
>>> path = default_storage.save("path/to/file", ContentFile(b"new content"))
>>> path
'path/to/file'
>>> default_storage.size(path)
11
>>> default_storage.open(path).read()
b'new content'
>>> default_storage.delete(path)
>>> default_storage.exists(path)
False
See :doc:`/ref/files/storage` for the file storage API.
.. _builtin-fs-storage:
The built-in filesystem storage class
-------------------------------------
Django ships with a :class:`django.core.files.storage.FileSystemStorage` class
which implements basic local filesystem file storage.
For example, the following code will store uploaded files under
``/media/photos`` regardless of what your :setting:`MEDIA_ROOT` setting is::
from django.core.files.storage import FileSystemStorage
from django.db import models
fs = FileSystemStorage(location="/media/photos")
class Car(models.Model):
...
photo = models.ImageField(storage=fs)
:doc:`Custom storage systems </howto/custom-file-storage>` work the same way:
you can pass them in as the ``storage`` argument to a
:class:`~django.db.models.FileField`.
Using a callable
----------------
You can use a callable as the :attr:`~django.db.models.FileField.storage`
parameter for :class:`~django.db.models.FileField` or
:class:`~django.db.models.ImageField`. This allows you to modify the used
storage at runtime, selecting different storages for different environments,
for example.
Your callable will be evaluated when your models classes are loaded, and must
return an instance of :class:`~django.core.files.storage.Storage`.
For example::
from django.conf import settings
from django.db import models
from .storages import MyLocalStorage, MyRemoteStorage
def select_storage():
return MyLocalStorage() if settings.DEBUG else MyRemoteStorage()
class MyModel(models.Model):
my_file = models.FileField(storage=select_storage)
In order to set a storage defined in the :setting:`STORAGES` setting you can
use :data:`~django.core.files.storage.storages`::
from django.core.files.storage import storages
def select_storage():
return storages["mystorage"]
class MyModel(models.Model):
upload = models.FileField(storage=select_storage)