I'm new using django. I've to upload a file and fot this I'm followin开发者_JAVA百科g the instructions present in official documentation: http://docs.djangoproject.com/en/dev/topics/http/file-uploads/?from=olddocs
my index.htlm <form action="upload_file" enctype="multipart/form-data" method="POST">
{% csrf_token %}
<input type="file" name="upfile" size="30">
<input type="submit" name="upfile" value= " Upload ">
</form>
my views.py:
def handle_uploaded_file(f):
destination = open('/my_path_to_tmp/tmp_files/input_file', 'wb+')
for chunk in f.chunks():
destination.write(chunk)
destination.close()
if ( f.file_name.endswith("sdf") ):
return "sdf"
elif ( f.file_name.endswith("smi") ):
return "smi"
def upload_file(request):
if request.method == "POST":
form = UploadFileForm(request.POST, request.FILES)
if form.is_valid():
file_type = handle_uploaded_file(request.FILES['upfile'])
return HttpResponseRedirect('calculate', file_type)
else:
form = UploadFileForm()
return render_to_response('upload.html', {'form': form})
my urls.py
urlpatterns = patterns('myapp.views',
(r'^upload_file$', 'upload_file'),
(r'calculate/$', 'calculation'),
)
Really I don't know what I'm doing wrong here but it seems that the condition
if request.method == "POST":
in views.py fails. Even if the method="POST" to the html form.
Anybody has an idea? Thank you so much!Are you sure that your form action is correct?
Shouldn't it be something like this instead:
<form action="{% url upload_file %}" enctype="multipart/form-data" method="post">
You could perhaps output the request.method at the beginning of your method, just to be sure... after that, print form._errors.
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