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