[Answered ]-Uploading files from a Python client to a Django Server

by

0👍

✅

It is actually working right now using Python’s requests module

Ill put the code for all interested…

Django server…

urls.py

...
url(r'^list/$', 'dataports.views.list', name='list'),
...

views.py

@csrf_exempt
def list(request):
    # Handle file upload
    if request.method == 'POST':
        print "upload file----------------------------------------------"
        form = DocumentForm(request.POST, request.FILES)
        if form.is_valid():
            print "otra vez.. es valido"
            print request.FILES

            newdoc = Jobpart(
                                partfile = request.FILES['docfile']
            )
            newdoc.save()

            # Redirect to the document list after POST
            return HttpResponseRedirect(reverse('dataports.views.list'))
    else:
        #print "nooooupload file----------------------------------------------"
        form = DocumentForm() # A empty, unbound form


    # Render list page with the documents and the form
    return render_to_response(
        'data_templates/list.html',
        {'form': form},
        context_instance=RequestContext(request)
    )

list.html

<!DOCTYPE html>
<html>
    <head>
        <meta charset="utf-8">
        <title>Minimal Django File Upload Example</title>   
    </head>

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

    </body>

</html>

Now in the client.

client.py

import requests
url = "http://localhost:8000/list/"
response = requests.post(url,files={'docfile': open('test.txt','rb')})

Now you are able to add some security and stuff.. But it is actually a very simple example..

Thank you all!!!!

👤ccamacho

2👍

What you want to do is to send a POST request to the Django app sending a file within it.

You can use python’s standard library httplib module or the 3rd party requests module.That last link posted shows how to post a multipart encoded file which is probably what you need.

Hope this helps!

👤Paulo Bu

1👍

Using requests:

with open('file') as f:
    requests.post('http://some.url/upload', data=f)
👤almalki

-1👍

file = request.FILES['file']

load_file = FileSystemStorage()

filename = load_file.save(file.name, file)  // saving in local directory and getting filename

data = {'name': name, 'address': address, 'age':age }

fr_data = None
with open(filepath ,'rb') as fr:
    fr_data += fr.read()

url = 'http://127.0.0.1:8000/api/'

response = requests.post(url=url, data=data, files= {
                          'filefiledname': fr_data
                          }
                         )
👤Kashyap Vijay

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