how to stream file to client in django - django

Question

I want to know how can I stream data to client using django.
The Goal
The user submits a form, the form data is passed to a web service which returns a string. The string is tarballed (tar.gz) and the tarball is sent back to the user.
I don't know what's the way. I searched and I found this, but I just have a string and I don't know if it is the thing I want, I don't know what to use in place of filename = __file__ , because I don't have file - just a string. If I create a new file for each user, this won't be a good way. so please help me. (sorry I'm new in web programming).
EDIT:
$('#sendButton').click(function(e) {
e.preventDefault();
var temp = $("#mainForm").serialize();
$.ajax({
type: "POST",
data: temp,
url: 'main/',
success: function(data) {
$("#mainDiv").html(data.form);
????
}
});
});
I want to use ajax, so what should i do in success of ajac function and in return of view. really thanks.
my view.py:
def idsBackup(request):
if request.is_ajax():
if request.method == 'POST':
result = ""
form = mainForm(request.POST)
if form.is_valid():
form = mainForm(request.POST)
//do form processing and call web service
string_to_return = webserviceString._result
???
to_json = {}
to_json['form'] = render_to_string('main.html', {'form': form}, context_instance=RequestContext(request))
to_json['result'] = result
???return HttpResponse(json.dumps(to_json), mimetype='application/json')
else:
form = mainForm()
return render_to_response('main.html', RequestContext(request, {'form':form}))
else:
return render_to_response("ajax.html", {}, context_instance=RequestContext(request))

Answer 1

You can create a django file instance of ContentFile using a string content instead of actual file and then send it as a response.
Sample code:
from django.core.files.base import ContentFile
def your_view(request):
#your view code
string_to_return = get_the_string() # get the string you want to return.
file_to_send = ContentFile(string_to_return)
response = HttpResponse(file_to_send,'application/x-gzip')
response['Content-Length'] = file_to_send.size
response['Content-Disposition'] = 'attachment; filename="somefile.tar.gz"'
return response

Answer 2

You can modify send_zipfile from the snippet to suit your needs. Just use StringIO to turn your string into a file-like object which can be passed to FileWrapper.
import StringIO, tempfile, zipfile
...
# get your string from the webservice
string = webservice.get_response()
...
temp = tempfile.TemporaryFile()
# this creates a zip, not a tarball
archive = zipfile.ZipFile(temp, 'w', zipfile.ZIP_DEFLATED)
# this converts your string into a filelike object
fstring = StringIO.StringIO(string)
# writes the "file" to the zip archive
archive.write(fstring)
archive.close()
wrapper = FileWrapper(temp)
response = HttpResponse(wrapper, content_type='application/zip')
response['Content-Disposition'] = 'attachment; filename=test.zip'
response['Content-Length'] = temp.tell()
temp.seek(0)
return response