How can I get the file name from request.FILES?

The answer may be outdated, since there is a name property on the UploadedFile class. See: Uploaded Files and Upload Handlers (Django docs). So, if you bind your form with a FileField correctly, the access should be as easy as:

if form.is_valid():
    form.cleaned_data['my_file'].name

request.FILES['filename'].name

From the request documentation.

If you don't know the key, you can iterate over the files:

for filename, file in request.FILES.iteritems():
    name = request.FILES[filename].name

file = request.FILES['filename']
file.name           # Gives name
file.content_type   # Gives Content type text/html etc
file.size           # Gives file's size in byte
file.read()         # Reads file

NOTE if you are using python 3.x:

request.FILES is a multivalue dictionary like object that keeps the files uploaded through an upload file button. Say in your html code the name of the button (type="file") is "myfile" so "myfile" will be the key in this dictionary. If you uploaded one file, then the value for this key will be only one and if you uploaded multiple files, then you will have multiple values for that specific key. If you use request.FILES['myfile'] you will get the first or last value (I cannot say for sure). This is fine if you only uploaded one file, but if you want to get all files you should do this:

list=[] #myfile is the key of a multi value dictionary, values are the uploaded files
for f in request.FILES.getlist('myfile'): #myfile is the name of your html file button
    filename = f.name
    list.append(filename)

of course one can squeeze the whole thing in one line, but this is easy to understand