Python - open uploaded file

Question

I have a Tornado web application where I want to read the an uploaded file. This is received from the client and I try to do so like this:

def post(self):

    file = self.request.files['images'][0]

    dataOpen = open(file['filename'],'r');
    dataRead = dataOpen.read()

But it gives an IOError:

Traceback (most recent call last):
  File "C:\Python27\lib\site-packages\tornado\web.py", line 1332, in _execute
    result = method(*self.path_args, **self.path_kwargs)
  File "C:\Users\rsaxdsxc\workspace\pi\src\Server.py", line 4100, in post
    dataOpen = open(file['filename'],'r');
  IOError: [Errno 2] No such file or directory: u'000c02c55024aeaa96e6c79bfa2de3926dbd3767.jpg'

Why isn't it able to see the file?

Well, you need to ensure that the directory matches where Python is trying to grab it from. Else you can use an absolute path. — Kyle Coventry, Feb 04 '15 at 19:10
Ensure that you write your file on upload before read it. With write(file['body']) — Most Wanted, Feb 04 '15 at 19:20

score 1 · Answer 1 · edited May 23 '17 at 12:28

1

Value of file['filename'] is just name of uploaded file, it is not path in your filesystem. Content of file is in file['body']. You can use StringIO module to emulate file interface if you want, or just directly iterate over file['body'].

Very good example you could use is here

So, your post request handler could look like:

def post(self):

    file = self.request.files['images'][0]
    dataRead = file['body']
    store_file_somewhere(file['filename'], dataRead)

edited May 23 '17 at 12:28

Community

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answered Feb 04 '15 at 19:28

Dmitry Nedbaylo

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Thanks, but if this an uploaded image that is going to be stored directly into a database then how would you create a path on a remote server to then save the file into the database? – user94628 Feb 04 '15 at 19:40
are you going to store path to image file in db or all the binary content of image file? – Dmitry Nedbaylo Feb 04 '15 at 20:52
I want to store straight into the DB and was able to do so with `GridFS` for mongoDB. Thanks – user94628 Feb 05 '15 at 11:25

Python - open uploaded file

1 Answers1