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File Uploads
.. currentmodule:: django.core.files.uploadedfile
When Django handles a file upload, the file data ends up placed in
:attr:`request.FILES <django.http.HttpRequest.FILES>` (for more on the
``request`` object see the documentation for :doc:`request and response objects
</ref/request-response>`). This document explains how files are stored on disk
and in memory, and how to customize the default behavior.
.. warning::
There are security risks if you are accepting uploaded content from
untrusted users! See the security guide's topic on
:ref:`user-uploaded-content-security` for mitigation details.
Basic file uploads
Consider a simple form containing a :class:`~django.forms.FileField`::
# In
from django import forms
class UploadFileForm(forms.Form):
title = forms.CharField(max_length=50)
file = forms.FileField()
A view handling this form will receive the file data in
:attr:`request.FILES <django.http.HttpRequest.FILES>`, which is a dictionary
containing a key for each :class:`~django.forms.FileField` (or
:class:`~django.forms.ImageField`, or other :class:`~django.forms.FileField`
subclass) in the form. So the data from the above form would
be accessible as ``request.FILES['file']``.
Note that :attr:`request.FILES <django.http.HttpRequest.FILES>` will only
contain data if the request method was ``POST`` and the ``<form>`` that posted
the request has the attribute ``enctype="multipart/form-data"``. Otherwise,
``request.FILES`` will be empty.
Most of the time, you'll simply pass the file data from ``request`` into the
form as described in :ref:`binding-uploaded-files`. This would look
something like::
from django.http import HttpResponseRedirect
from django.shortcuts import render_to_response
from .forms import UploadFileForm
# Imaginary function to handle an uploaded file.
from somewhere import handle_uploaded_file
def upload_file(request):
if request.method == 'POST':
form = UploadFileForm(request.POST, request.FILES)
if form.is_valid():
return HttpResponseRedirect('/success/url/')
form = UploadFileForm()
return render_to_response('upload.html', {'form': form})
Notice that we have to pass :attr:`request.FILES <django.http.HttpRequest.FILES>`
into the form's constructor; this is how file data gets bound into a form.
Here's a common way you might handle an uploaded file::
def handle_uploaded_file(f):
with open('some/file/name.txt', 'wb+') as destination:
for chunk in f.chunks():
Looping over ``UploadedFile.chunks()`` instead of using ``read()`` ensures that
large files don't overwhelm your system's memory.
There are a few other methods and attributes available on ``UploadedFile``
objects; see :class:`UploadedFile` for a complete reference.
Handling uploaded files with a model
If you're saving a file on a :class:`~django.db.models.Model` with a
:class:`~django.db.models.FileField`, using a :class:`~django.forms.ModelForm`
makes this process much easier. The file object will be saved to the location
specified by the :attr:`~django.db.models.FileField.upload_to` argument of the
corresponding :class:`~django.db.models.FileField` when calling
from django.http import HttpResponseRedirect
from django.shortcuts import render
from .forms import ModelFormWithFileField
def upload_file(request):
if request.method == 'POST':
form = ModelFormWithFileField(request.POST, request.FILES)
if form.is_valid():
# file is saved
return HttpResponseRedirect('/success/url/')
form = ModelFormWithFileField()
return render(request, 'upload.html', {'form': form})
If you are constructing an object manually, you can simply assign the file
object from :attr:`request.FILES <django.http.HttpRequest.FILES>` to the file
field in the model::
from django.http import HttpResponseRedirect
from django.shortcuts import render
from .forms import UploadFileForm
from .models import ModelWithFileField
def upload_file(request):
if request.method == 'POST':
form = UploadFileForm(request.POST, request.FILES)
if form.is_valid():
instance = ModelWithFileField(file_field=request.FILES['file'])
return HttpResponseRedirect('/success/url/')
form = UploadFileForm()
return render(request, 'upload.html', {'form': form})
Upload Handlers
.. currentmodule:: django.core.files.uploadhandler
When a user uploads a file, Django passes off the file data to an *upload
handler* -- a small class that handles file data as it gets uploaded. Upload
handlers are initially defined in the :setting:`FILE_UPLOAD_HANDLERS` setting,
which defaults to::
Together :class:`MemoryFileUploadHandler` and
:class:`TemporaryFileUploadHandler` provide Django's default file upload
behavior of reading small files into memory and large ones onto disk.
You can write custom handlers that customize how Django handles files. You
could, for example, use custom handlers to enforce user-level quotas, compress
data on the fly, render progress bars, and even send data to another storage
location directly without storing it locally. See :ref:`custom_upload_handlers`
for details on how you can customize or completely replace upload behavior.
.. _modifying_upload_handlers_on_the_fly:
Where uploaded data is stored
Before you save uploaded files, the data needs to be stored somewhere.
By default, if an uploaded file is smaller than 2.5 megabytes, Django will hold
the entire contents of the upload in memory. This means that saving the file
involves only a read from memory and a write to disk and thus is very fast.
However, if an uploaded file is too large, Django will write the uploaded file
to a temporary file stored in your system's temporary directory. On a Unix-like
platform this means you can expect Django to generate a file called something
like ``/tmp/tmpzfp6I6.upload``. If an upload is large enough, you can watch this
file grow in size as Django streams the data onto disk.
These specifics -- 2.5 megabytes; ``/tmp``; etc. -- are simply "reasonable
defaults" which can be customized as described in the next section.
Changing upload handler behavior
There are a few settings which control Django's file upload behavior. See
:ref:`File Upload Settings <file-upload-settings>` for details.
Modifying upload handlers on the fly
Sometimes particular views require different upload behavior. In these cases,
you can override upload handlers on a per-request basis by modifying
``request.upload_handlers``. By default, this list will contain the upload
handlers given by :setting:`FILE_UPLOAD_HANDLERS`, but you can modify the list
as you would any other list.
For instance, suppose you've written a ``ProgressBarUploadHandler`` that
provides feedback on upload progress to some sort of AJAX widget. You'd add this
handler to your upload handlers like this::
request.upload_handlers.insert(0, ProgressBarUploadHandler())
You'd probably want to use ``list.insert()`` in this case (instead of
``append()``) because a progress bar handler would need to run *before* any
other handlers. Remember, the upload handlers are processed in order.
If you want to replace the upload handlers completely, you can just assign a new
request.upload_handlers = [ProgressBarUploadHandler()]
.. note::
You can only modify upload handlers *before* accessing
``request.POST`` or ``request.FILES`` -- it doesn't make sense to
change upload handlers after upload handling has already
started. If you try to modify ``request.upload_handlers`` after
reading from ``request.POST`` or ``request.FILES`` Django will
throw an error.
Thus, you should always modify uploading handlers as early in your view as
Also, ``request.POST`` is accessed by
:class:`~django.middleware.csrf.CsrfViewMiddleware` which is enabled by
default. This means you will need to use
:func:`~django.views.decorators.csrf.csrf_exempt` on your view to allow you
to change the upload handlers. You will then need to use
:func:`~django.views.decorators.csrf.csrf_protect` on the function that
actually processes the request. Note that this means that the handlers may
start receiving the file upload before the CSRF checks have been done.
Example code::
from django.views.decorators.csrf import csrf_exempt, csrf_protect
def upload_file_view(request):
request.upload_handlers.insert(0, ProgressBarUploadHandler())
return _upload_file_view(request)
def _upload_file_view(request):
... # Process request
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