django forms without model

Question

I want to upload an image from a form without model. I've tried to make:

In the template: det.html

Upload image
    <form enctype="multipart/form-data" action="" method="post">{% csrf_token %}
        <input type="file" name="myfile" id="myfile" />
        <input type="submit" value="Upload image">
    </form>

In the views, I have:

def fileupload(request):

    return responsewrapper('personne/det.html', locals(),request)

def handle_uploaded_file(f):
    filename = file._get_name()
    destination = open('/personne/static/personne/%s'%filename, 'wb+')
    for chunk in f.chunks():
        destination.write(chunk)
    destination.close()

def submitfilecontent(request):
    handle_uploaded_file(request.FILES['myfile'])
    return HttpResponseRedirect("/successupload")

The error is: Method Not Allowed (POST): /fr/detect/

my urls:

urlpatterns += i18n_patterns(
    url(r'^admin/', admin.site.urls, name="admin"),
    url(r'^$', views.IndexView.as_view(), name="homepage"),
    url(r'^detect/$', views.DetectView.as_view(), name="detection"),
    url(r'^login/$', auth_views.login, name="login"),
    url(r'^logout/$', auth_views.logout, name="logout"),
    url(r'^register/',views.addUser, name='register'),

)+ i18n_patterns('', (r'^(?P<path>.*)$', 'django.views.static.serve', {'document_root': MEDIA_ROOT, 'show_indexes': True}))

Update:

in urls.py, I added:

  url(r'^submitfilecontent/$', submitfilecontent, name="submit-file-content"),

in the template:

<form enctype="multipart/form-data" action="/submitfilecontent/" method="post">

The new error: MultiValueDictKeyError: "'myfile'"

I think the problem seems to be authentication issue. somthing like CORS or csrf issue or a simple URL without a slash. Can you show the url in urls.py — SuperNova, Apr 10 '16 at 11:06
Also, the action attribute in form tag does not have the url. — SuperNova, Apr 10 '16 at 11:07
The action url should be the one which you intend to use to handle upload. — SuperNova, Apr 10 '16 at 11:14

score 1 · Answer 1 · answered Apr 10 '16 at 11:17

1

Please add a url(below) in urls.

url(r'^submitfilecontent/$', submitfilecontent, name="submit-file-content"),

and in HTML, add the action url

<form enctype="multipart/form-data" action="/submitfilecontent/" method="post">{% csrf_token %}

answered Apr 10 '16 at 11:17

SuperNova

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I'm sorry for my lack of knowledge, I have now this error: handle_uploaded_file(request.FILES['myfile']) # error throws up here. – Lilia Apr 10 '16 at 11:31
MultiValueDictKeyError: "'myfile'" – Lilia Apr 10 '16 at 11:44
No, I used this link to begin: http://stackoverflow.com/questions/21882298/django-uploading-files-without-model – Lilia Apr 10 '16 at 11:50
Plzz visit these links & follow the steps. http://www.bogotobogo.com/python/Django/Python_Django_Image_Files_Uploading_Example.php & http://stackoverflow.com/questions/5871730/need-a-minimal-django-file-upload-example – SuperNova Apr 10 '16 at 11:56

score 0 · Accepted Answer · answered Apr 10 '16 at 15:10

Views.py:

def submitfilecontent(request):
    ext_allowed = ['gif', 'jpg', 'jpeg', 'png']
    today = datetime.datetime.today()
    save_dir = 'personne/static/personne/%d/%d/%d/' % (today.year, today.month, today.day)
    save_path = os.path.join(settings.MEDIA_ROOT, save_dir)
    save_url = os.path.join(settings.MEDIA_URL, save_dir)

    if request.method == 'POST':
        file = request.FILES['myfile']
        ext = file.name.split('.').pop()
        if not os.path.isdir(save_path):
            os.makedirs(save_path)

        new_file = '%s.%s' % (int(time.time()), ext)

        destination = open(save_path+new_file, 'wb+')
        for chunk in file.chunks():
            destination.write(chunk)
        destination.close()
        return  HttpResponse("Upload Succsefull to URL:%s" % (save_url+new_file)  )
    else:
        raise Http404

django forms without model

2 Answers2