As a newbie to Django, I am having difficulty making an upload app in Django 1.3. I could not find any up-to-date example/snippets. May someone post a minimal but complete (Model, View, Template) example code to do so?

Solution 1

Phew, Django documentation really does not have good example about this. I spent over 2 hours to dig up all the pieces to understand how this works. With that knowledge I implemented a project that makes possible to upload files and show them as list. To download source for the project, visit or clone it:

> git clone

Update 2013-01-30: The source at GitHub has also implementation for Django 1.4 in addition to 1.3. Even though there is few changes the following tutorial is also useful for 1.4.

Update 2013-05-10: Implementation for Django 1.5 at GitHub. Minor changes in redirection in and usage of url template tag in list.html. Thanks to hubert3 for the effort.

Update 2013-12-07: Django 1.6 supported at GitHub. One import changed in myapp/ Thanks goes to Arthedian.

Update 2015-03-17: Django 1.7 supported at GitHub, thanks to aronysidoro.

Update 2015-09-04: Django 1.8 supported at GitHub, thanks to nerogit.

Update 2016-07-03: Django 1.9 supported at GitHub, thanks to daavve and nerogit

Project tree

A basic Django 1.3 project with single app and media/ directory for uploads.


1. Settings: myproject/

To upload and serve files, you need to specify where Django stores uploaded files and from what URL Django serves them. MEDIA_ROOT and MEDIA_URL are in by default but they are empty. See the first lines in Django Managing Files for details. Remember also set the database and add myapp to INSTALLED_APPS

import os

BASE_DIR = os.path.dirname(os.path.dirname(__file__))
    'default': {
        'ENGINE': 'django.db.backends.sqlite3',
        'NAME': os.path.join(BASE_DIR, 'database.sqlite3'),
        'USER': '',
        'PASSWORD': '',
        'HOST': '',
        'PORT': '',
MEDIA_ROOT = os.path.join(BASE_DIR, 'media')
MEDIA_URL = '/media/'

2. Model: myproject/myapp/

Next you need a model with a FileField. This particular field stores files e.g. to media/documents/2011/12/24/ based on current date and MEDIA_ROOT. See FileField reference.

# -*- coding: utf-8 -*-
from django.db import models

class Document(models.Model):
    docfile = models.FileField(upload_to='documents/%Y/%m/%d')

3. Form: myproject/myapp/

To handle upload nicely, you need a form. This form has only one field but that is enough. See Form FileField reference for details.

# -*- coding: utf-8 -*-
from django import forms

class DocumentForm(forms.Form):
    docfile = forms.FileField(
        label='Select a file',
        help_text='max. 42 megabytes'

4. View: myproject/myapp/

A view where all the magic happens. Pay attention how request.FILES are handled. For me, it was really hard to spot the fact that request.FILES['docfile'] can be saved to models.FileField just like that. The model's save() handles the storing of the file to the filesystem automatically.

# -*- coding: utf-8 -*-
from django.shortcuts import render_to_response
from django.template import RequestContext
from django.http import HttpResponseRedirect
from django.core.urlresolvers import reverse

from myproject.myapp.models import Document
from myproject.myapp.forms import DocumentForm

def list(request):
    # Handle file upload
    if request.method == 'POST':
        form = DocumentForm(request.POST, request.FILES)
        if form.is_valid():
            newdoc = Document(docfile = request.FILES['docfile'])

            # Redirect to the document list after POST
            return HttpResponseRedirect(reverse('myapp.views.list'))
        form = DocumentForm() # A empty, unbound form

    # Load documents for the list page
    documents = Document.objects.all()

    # Render list page with the documents and the form
    return render_to_response(
        {'documents': documents, 'form': form},

5. Project URLs: myproject/

Django does not serve MEDIA_ROOT by default. That would be dangerous in production environment. But in development stage, we could cut short. Pay attention to the last line. That line enables Django to serve files from MEDIA_URL. This works only in developement stage.

See django.conf.urls.static.static reference for details. See also this discussion about serving media files.

# -*- coding: utf-8 -*-
from django.conf.urls import patterns, include, url
from django.conf import settings
from django.conf.urls.static import static

urlpatterns = patterns('',
    (r'^', include('myapp.urls')),
) + static(settings.MEDIA_URL, document_root=settings.MEDIA_ROOT)

6. App URLs: myproject/myapp/

To make the view accessible, you must specify urls for it. Nothing special here.

# -*- coding: utf-8 -*-
from django.conf.urls import patterns, url

urlpatterns = patterns('myapp.views',
    url(r'^list/$', 'list', name='list'),

7. Template: myproject/myapp/templates/myapp/list.html

The last part: template for the list and the upload form below it. The form must have enctype-attribute set to "multipart/form-data" and method set to "post" to make upload to Django possible. See File Uploads documentation for details.

The FileField has many attributes that can be used in templates. E.g. {{ document.docfile.url }} and {{ }} as in the template. See more about these in Using files in models article and The File object documentation.

<!DOCTYPE html>
        <meta charset="utf-8">
        <title>Minimal Django File Upload Example</title>   
    <!-- List of uploaded documents -->
    {% if documents %}
        {% for document in documents %}
            <li><a href="{{ document.docfile.url }}">{{ }}</a></li>
        {% endfor %}
    {% else %}
        <p>No documents.</p>
    {% endif %}

        <!-- Upload form. Note enctype attribute! -->
        <form action="{% url 'list' %}" method="post" enctype="multipart/form-data">
            {% csrf_token %}
            <p>{{ form.non_field_errors }}</p>
            <p>{{ form.docfile.label_tag }} {{ form.docfile.help_text }}</p>
                {{ form.docfile.errors }}
                {{ form.docfile }}
            <p><input type="submit" value="Upload" /></p>

8. Initialize

Just run syncdb and runserver.

> cd myproject
> python syncdb
> python runserver


Finally, everything is ready. On default Django developement environment the list of uploaded documents can be seen at localhost:8000/list/. Today the files are uploaded to /path/to/myproject/media/documents/2011/12/17/ and can be opened from the list.

I hope this answer will help someone as much as it would have helped me.

Solution 2


See the github repo, works with Django 3

A minimal Django file upload example

1. Create a django project

Run startproject::

$ startproject sample

now a folder(sample) is created.

2. create an app

Create an app::

$ cd sample
$ python startapp uploader

Now a folder(uploader) with these files are created::


3. Update

On sample/ add 'uploader' to INSTALLED_APPS and add MEDIA_ROOT and MEDIA_URL, ie::

    ...<other apps>...      

MEDIA_ROOT = os.path.join(BASE_DIR, 'media')
MEDIA_URL = '/media/'

4. Update

in sample/ add::

...<other imports>...
from django.conf import settings
from django.conf.urls.static import static
from uploader import views as uploader_views

urlpatterns = [
    ...<other url patterns>...
    path('', uploader_views.UploadView.as_view(), name='fileupload'),
]+ static(settings.MEDIA_URL, document_root=settings.MEDIA_ROOT)

5. Update

update uploader/

from django.db import models
class Upload(models.Model):
    upload_file = models.FileField()    
    upload_date = models.DateTimeField(auto_now_add =True)

6. Update

update uploader/

from django.views.generic.edit import CreateView
from django.urls import reverse_lazy
from .models import Upload
class UploadView(CreateView):
    model = Upload
    fields = ['upload_file', ]
    success_url = reverse_lazy('fileupload')
    def get_context_data(self, **kwargs):
        context = super().get_context_data(**kwargs)
        context['documents'] = Upload.objects.all()
        return context

7. create templates

Create a folder sample/uploader/templates/uploader

Create a file upload_form.html ie sample/uploader/templates/uploader/upload_form.html::

<div style="padding:40px;margin:40px;border:1px solid #ccc">
    <h1>Django File Upload</h1>
    <form method="post" enctype="multipart/form-data">
      {% csrf_token %}
      {{ form.as_p }}
      <button type="submit">Submit</button>
    {% for document in documents %}
            <a href="{{ document.upload_file.url }}">{{ }}</a>
            <small>({{ document.upload_file.size|filesizeformat }}) - {{document.upload_date}}</small>
    {% endfor %}

8. Syncronize database

Syncronize database and runserver::

$ python makemigrations
$ python migrate
$ python runserver

visit http://localhost:8000/

Solution 3

Generally speaking when you are trying to 'just get a working example' it is best to 'just start writing code'. There is no code here to help you with, so it makes answering the question a lot more work for us.

If you want to grab a file, you need something like this in an html file somewhere:

<form method="post" enctype="multipart/form-data">
    <input type="file" name="myfile" />
    <input type="submit" name="submit" value="Upload" />

That will give you the browse button, an upload button to start the action (submit the form) and note the enctype so Django knows to give you request.FILES

In a view somewhere you can access the file with

def myview(request):
    request.FILES['myfile'] # this is my file

There is a huge amount of information in the file upload docs

I recommend you read the page thoroughly and just start writing code - then come back with examples and stack traces when it doesn't work.

Solution 4

I must say I find the documentation at django confusing. Also for the simplest example why are forms being mentioned? The example I got to work in the is :-

for key, file in request.FILES.items():
    path =
    dest = open(path, 'w')
    if file.multiple_chunks:
        for c in file.chunks():

The html file looks like the code below, though this example only uploads one file and the code to save the files handles many :-

<form action="/upload_file/" method="post" enctype="multipart/form-data">{% csrf_token %}
<label for="file">Filename:</label>
<input type="file" name="file" id="file" />
<br />
<input type="submit" name="submit" value="Submit" />

These examples are not my code, they have been optained from two other examples I found. I am a relative beginner to django so it is very likely I am missing some key point.

Solution 5

I also had the similar requirement. Most of the examples on net are asking to create models and create forms which I did not wanna use. Here is my final code.

if request.method == 'POST':
    file1 = request.FILES['file']
    contentOfFile =
    if file1:
        return render(request, 'blogapp/Statistics.html', {'file': file1, 'contentOfFile': contentOfFile})

And in HTML to upload I wrote:

{% block content %}
    <h1>File content</h1>
    <form action="{% url 'blogapp:uploadComplete'%}" method="post" enctype="multipart/form-data">
         {% csrf_token %}
        <input id="uploadbutton" type="file" value="Browse" name="file" accept="text/csv" />
        <input type="submit" value="Upload" />
    {% endblock %}

Following is the HTML which displays content of file:

{% block content %}
    <h3>File uploaded successfully</h3>
    </br>content = {{contentOfFile}}
{% endblock %}

Solution 6

Extending on Henry's example:

import tempfile
import shutil

FILE_UPLOAD_DIR = '/home/imran/uploads'

def handle_uploaded_file(source):
    fd, filepath = tempfile.mkstemp(, dir=FILE_UPLOAD_DIR)
    with open(filepath, 'wb') as dest:
        shutil.copyfileobj(source, dest)
    return filepath

You can call this handle_uploaded_file function from your view with the uploaded file object. This will save the file with a unique name (prefixed with filename of the original uploaded file) in filesystem and return the full path of saved file. You can save the path in database, and do something with the file later.

Solution 7

Here it may helps you: create a file field in your

For uploading the file(in your

def save_model(self, request, obj, form, change):
    url = "" %(
    url = str(url)

    if url:
        temp_img = NamedTemporaryFile(delete=True)
        filename_img = urlparse(url).path.split('/')[-1],File(temp_img)

and use that field in your template also.

Solution 8

You can refer to server examples in Fine Uploader, which has django version.

It's very elegant and most important of all, it provides featured js lib. Template is not included in server-examples, but you can find demo on its website. Fine Uploader:


UploadView dispatches post and delete request to respective handlers.

class UploadView(View):

    def dispatch(self, *args, **kwargs):
        return super(UploadView, self).dispatch(*args, **kwargs)

    def post(self, request, *args, **kwargs):
        """A POST request. Validate the form and then handle the upload
        based ont the POSTed data. Does not handle extra parameters yet.
        form = UploadFileForm(request.POST, request.FILES)
        if form.is_valid():
            handle_upload(request.FILES['qqfile'], form.cleaned_data)
            return make_response(content=json.dumps({ 'success': True }))
            return make_response(status=400,
                    'success': False,
                    'error': '%s' % repr(form.errors)

    def delete(self, request, *args, **kwargs):
        """A DELETE request. If found, deletes a file with the corresponding
        UUID from the server's filesystem.
        qquuid = kwargs.get('qquuid', '')
        if qquuid:
                return make_response(content=json.dumps({ 'success': True }))
            except Exception, e:
                return make_response(status=400,
                        'success': False,
                        'error': '%s' % repr(e)
        return make_response(status=404,
                'success': False,
                'error': 'File not present'

class UploadFileForm(forms.Form):

    """ This form represents a basic request from Fine Uploader.
    The required fields will **always** be sent, the other fields are optional
    based on your setup.
    Edit this if you want to add custom parameters in the body of the POST
    qqfile = forms.FileField()
    qquuid = forms.CharField()
    qqfilename = forms.CharField()
    qqpartindex = forms.IntegerField(required=False)
    qqchunksize = forms.IntegerField(required=False)
    qqpartbyteoffset = forms.IntegerField(required=False)
    qqtotalfilesize = forms.IntegerField(required=False)
    qqtotalparts = forms.IntegerField(required=False)

Solution 9

Not sure if there any disadvantages to this approach but even more minimal, in

entry =

# save uploaded file
if request.FILES['myfile']:['myfile']._name, request.FILES['myfile'], True)

Solution 10

I faced the similar problem, and solved by django admin site.

# models
class Document(models.Model):
    docfile = models.FileField(upload_to='documents/Temp/%Y/%m/%d')

    def doc_name(self):
        return'/')[-1] # only the name, not full path

# admin
from myapp.models import Document
class DocumentAdmin(admin.ModelAdmin):
    list_display = ('doc_name',), DocumentAdmin)