How to upload a file in Django

Question

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

User · Answer

You can refer to server examples in Fine Uploader, which has django version. https://github.com/FineUploader/server-examples/tree/master/python/django-fine-uploader

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: http://fineuploader.com/demos.html

django-fine-uploader

views.py

UploadView dispatches post and delete request to respective handlers.

class UploadView(View):

    @csrf_exempt
    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 }))
        else:
            return make_response(status=400,
                content=json.dumps({
                    '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:
            try:
                handle_deleted_file(qquuid)
                return make_response(content=json.dumps({ 'success': True }))
            except Exception, e:
                return make_response(status=400,
                    content=json.dumps({
                        'success': False,
                        'error': '%s' % repr(e)
                    }))
        return make_response(status=404,
            content=json.dumps({
                'success': False,
                'error': 'File not present'
            }))

forms.py

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
    request.
    """
    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)

User · Answer

Demo  See the github repo  works with Django 3  A minimal Django file upload example  1  Create a django project  Run startproject      django-admin py startproject sample   now a folder sample  is created   2  create an app  Create an app      cd sample   python manage py startapp uploader   Now a folder uploader  with these files are created    uploader      init   py   admin py   app py   models py   tests py   views py   migrations        init   py   3  Update settings py  On sample settings py add  uploader  to INSTALLED APPS and add MEDIA ROOT and MEDIA URL  ie    INSTALLED APPS          uploader           lt other apps gt              MEDIA ROOT   os path join BASE DIR   media   MEDIA URL     media     4  Update urls py  in sample urls py add        lt other imports gt     from django conf import settings from django conf urls static import static from uploader import views as uploader views  urlpatterns             lt other url patterns gt         path     uploader views UploadView as view    name  fileupload       static settings MEDIA URL  document root settings MEDIA ROOT    5  Update models py  update uploader models py    from django db import models class Upload models Model       upload file   models FileField           upload date   models DateTimeField auto now add  True    6  Update views py  update uploader views py    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     lt div style  padding 40px margin 40px border 1px solid  ccc  gt       lt h1 gt Django File Upload lt  h1 gt       lt form method  post  enctype  multipart form-data  gt           csrf token             form as p           lt button type  submit  gt Submit lt  button gt       lt  form gt  lt hr gt       lt ul gt         for document in documents             lt li gt               lt a href     document upload file url     gt    document upload file name    lt  a gt               lt small gt     document upload file size filesizeformat     -   document upload date   lt  small gt           lt  li gt         endfor         lt  ul gt   lt  div gt    8  Syncronize database  Syncronize database and runserver      python manage py makemigrations   python manage py migrate   python manage py runserver   visit http   localhost 8000

User · Answer

Extending on Henry s example   import tempfile import shutil  FILE UPLOAD DIR     home imran uploads   def handle uploaded file source       fd  filepath   tempfile mkstemp prefix source name  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

User · Answer

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   file1 read       if file1          return render request   blogapp Statistics html     file   file1   contentOfFile   contentOfFile     And in HTML to upload I wrote      block content         lt h1 gt File content lt  h1 gt       lt form action     url  blogapp uploadComplete     method  post  enctype  multipart form-data  gt              csrf token             lt input id  uploadbutton  type  file  value  Browse  name  file  accept  text csv    gt           lt input type  submit  value  Upload    gt       lt  form gt         endblock      Following is the HTML which displays content of file      block content         lt h3 gt File uploaded successfully lt  h3 gt        file name        lt  br gt content     contentOfFile      endblock

User · Answer

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 views py is  -  for key  file in request FILES items        path   file name     dest   open path   w       if file multiple chunks          for c in file chunks                dest write c      else          dest write file read        dest close     The html file looks like the code below  though this example only uploads one file and the code to save the files handles many  -   lt form action   upload file   method  post  enctype  multipart form-data  gt    csrf token     lt label for  file  gt Filename  lt  label gt   lt input type  file  name  file  id  file    gt   lt br   gt   lt input type  submit  name  submit  value  Submit    gt   lt  form gt    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

User · Answer

Here it may helps you   create a file field in your models py  For uploading the file in your admin py    def save model self  request  obj  form  change       url    http   img youtube com vi  s hqdefault jpg    obj video      url   str url       if url          temp img   NamedTemporaryFile delete True          temp img write urllib2 urlopen url  read            temp img flush           filename img   urlparse url  path split      -1          obj image save filename img File temp img    and use that field in your template also

User · Answer

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 https   github com axelpale minimal-django-file-upload-example or clone it    gt  git clone https   github com axelpale minimal-django-file-upload-example git   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 urls py 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 urls py  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    minimal-django-file-upload-example      src          myproject              database                  sqlite db             media              myapp                  templates                      myapp                          list html                 forms py                 models py                 urls py                 views py               init   py             manage py             settings py             urls py   1  Settings  myproject settings py  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 settings py 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         DATABASES          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       INSTALLED APPS                  myapp       2  Model  myproject myapp models py  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 forms py  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 views py  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                newdoc save                  Redirect to the document list after POST             return HttpResponseRedirect reverse  myapp views list        else          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           myapp list html             documents   documents   form   form           context instance RequestContext request          5  Project URLs  myproject urls py  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 urls py  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    document docfile name    as in the template  See more about these in Using files in models article and The File object documentation    lt  DOCTYPE html gt   lt html gt       lt head gt           lt meta charset  utf-8  gt           lt title gt Minimal Django File Upload Example lt  title gt          lt  head gt       lt body gt       lt  -- List of uploaded documents -- gt         if documents             lt ul gt             for document in documents                 lt li gt  lt a href     document docfile url     gt    document docfile name    lt  a gt  lt  li gt             endfor             lt  ul gt         else             lt p gt No documents  lt  p gt         endif              lt  -- Upload form  Note enctype attribute  -- gt           lt form action     url  list      method  post  enctype  multipart form-data  gt                 csrf token                 lt p gt    form non field errors    lt  p gt               lt p gt    form docfile label tag       form docfile help text    lt  p gt               lt p gt                     form docfile errors                       form docfile                 lt  p gt               lt p gt  lt input type  submit  value  Upload    gt  lt  p gt           lt  form gt       lt  body gt   lt  html gt     8  Initialize  Just run syncdb and runserver    gt  cd myproject  gt  python manage py syncdb  gt  python manage py runserver   Results  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

User · Answer

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    lt form method  post  enctype  multipart form-data  gt       lt input type  file  name  myfile    gt       lt input type  submit  name  submit  value  Upload    gt   lt  form gt    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

User · Answer

Not sure if there any disadvantages to this approach but even more minimal  in views py   entry   form save      save uploaded file if request FILES  myfile        entry myfile save request FILES  myfile    name  request FILES  myfile    True

User · Answer

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 self docfile name split      -1    only the name  not full path    admin from myapp models import Document class DocumentAdmin admin ModelAdmin       list display     doc name    admin site register Document  DocumentAdmin

[django] How to upload a file in Django?

django-fine-uploader

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