django/docs/topics/http/file-uploads.txt

============
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 form containing a :class:`~django.forms.FileField`:

.. code-block:: python
    :caption: ``forms.py``

    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``, at least one file field was
actually posted, 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 pass the file data from ``request`` into the form as
described in :ref:`binding-uploaded-files`. This would look something like:

.. code-block:: python
    :caption: ``views.py``

    from django.http import HttpResponseRedirect
    from django.shortcuts import render
    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():
                handle_uploaded_file(request.FILES['file'])
                return HttpResponseRedirect('/success/url/')
        else:
            form = UploadFileForm()
        return render(request, '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():
                destination.write(chunk)

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
``form.save()``::

    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
                form.save()
                return HttpResponseRedirect('/success/url/')
        else:
            form = ModelFormWithFileField()
        return render(request, 'upload.html', {'form': form})

If you are constructing an object manually, you can 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'])
                instance.save()
                return HttpResponseRedirect('/success/url/')
        else:
            form = UploadFileForm()
        return render(request, 'upload.html', {'form': form})

If you are constructing an object manually outside of a request, you can assign
a :class:`~django.core.files.File` like object to the
:class:`~django.db.models.FileField`::

    from django.core.management.base import BaseCommand
    from django.core.files.base import ContentFile

    class MyCommand(BaseCommand):
        def handle(self, *args, **options):
            content_file = ContentFile(b'Hello world!', name='hello-world.txt')
            instance = ModelWithFileField(file_field=content_file)
            instance.save()

Uploading multiple files
------------------------

If you want to upload multiple files using one form field, set the ``multiple``
HTML attribute of field's widget:

.. code-block:: python
    :caption: ``forms.py``

    from django import forms

    class FileFieldForm(forms.Form):
        file_field = forms.FileField(widget=forms.ClearableFileInput(attrs={'multiple': True}))

Then override the ``post`` method of your
:class:`~django.views.generic.edit.FormView` subclass to handle multiple file
uploads:

.. code-block:: python
    :caption: ``views.py``

    from django.views.generic.edit import FormView
    from .forms import FileFieldForm

    class FileFieldFormView(FormView):
        form_class = FileFieldForm
        template_name = 'upload.html'  # Replace with your template.
        success_url = '...'  # Replace with your URL or reverse().

        def post(self, request, *args, **kwargs):
            form_class = self.get_form_class()
            form = self.get_form(form_class)
            files = request.FILES.getlist('file_field')
            if form.is_valid():
                for f in files:
                    ...  # Do something with each file.
                return self.form_valid(form)
            else:
                return self.form_invalid(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::

    ["django.core.files.uploadhandler.MemoryFileUploadHandler",
     "django.core.files.uploadhandler.TemporaryFileUploadHandler"]

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.

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 "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:

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(request))

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 assign a new
list::

   request.upload_handlers = [ProgressBarUploadHandler(request)]

.. 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
    possible.

    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

        @csrf_exempt
        def upload_file_view(request):
            request.upload_handlers.insert(0, ProgressBarUploadHandler(request))
            return _upload_file_view(request)

        @csrf_protect
        def _upload_file_view(request):
            ... # Process request

    If you are using a class-based view, you will need to use
    :func:`~django.views.decorators.csrf.csrf_exempt` on its
    :meth:`~django.views.generic.base.View.dispatch` method and
    :func:`~django.views.decorators.csrf.csrf_protect` on the method that
    actually processes the request. Example code::

        from django.utils.decorators import method_decorator
        from django.views import View
        from django.views.decorators.csrf import csrf_exempt, csrf_protect

        @method_decorator(csrf_exempt, name='dispatch')
        class UploadFileView(View):

            def setup(self, request, *args, **kwargs):
                request.upload_handlers.insert(0, ProgressBarUploadHandler(request))
                super().setup(request, *args, **kwargs)

            @method_decorator(csrf_protect)
            def post(self, request, *args, **kwargs):
                ... # Process request