Question

Firstly – sorry for my English..
I’ve inherited a rather big project which uses ExtJS and is based on Django.. So I have an upload form and it works ok. The path to the file is given in a fileuploadfield (this is the xtype of ExtJS field) then it is taken by the Django and saved to the DB (it’s being transformed to a needed view to be stored in the base, but that’s not the problem..)
So I have smth like this:

class image(entity):
    ...

    def create(self,request):
    if request.method == 'POST':
        form = UploadFileForm(request.POST, request.FILES)
        try:
            parent_id = int(request.POST["parent_id"])
        except:
            return ErrorResponse()
        if form.is_valid():
            fl =  request.FILES['file'];
            file = fl.read()
            Obj = self.data.obj
            New = Obj()

            setattr(New,'is_main',False)

and so on..
So I need to make a file upload from a direct link. Of course by another field. Having the file from a link is not a problem (urllib2 helps great), saving the file to the media folder isn’t a problem too. But I can’t insert it in the code above. As I understand – my problem is that the file object differs from the request.FILES thing.. But I don’t know how to convert a normal file (in my case all the files are images) to the type of request.FILES so it would be readable for the next functions.. What I mean:
I catch the link to the file from another field(textfield):

            url = request.POST['link'] #It get's the user posted link
            name = urllib2.urlopen(url).read()
            f = open('media/images/picture.gif', 'wb') #I overwrite it on a dummy, couse it must be than transfered to the db
            f.write(name)
            f.close #the file can be seen in the media path

Ok. The file is now in the /media/ folder, but if I simply try to replace the
file = fl.read() with file=f.read() it doesn’t work. As I understand – the request.FILES and just a file are different.. Can anybody help me to convert my file to the needed type? Or tell me how can this idea be done.. The project is really big and if I try changing the submitting mechanism it may cause a lot of problems..
Again sorry for my English)

Answer 1

I really don’t understand why it didn’t work, but I found out that this works for me..

            url = request.POST['file_link']
            name = urllib2.urlopen(url).read()
            file = name
            ..

the urllib2.urlopen(“link to a file”) returns a normal File and that is it..