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I try uploading a file using the following code (which has been summarized).

On the server-side, when I check the name of the received file, I see that it is set to the file_path parameter.

Currently, the poster library takes the file_path parameter that I entered (e.g.: file_path=~/user/data.csv), and sends it as the file’s name during the file multipart uploading.

I am looking for a way to change the file's name that is being sent (Since the server expects a file name without the / char)

Thanks in advance.

P.S. I have already tried adding the "name" key to the values, but it didn't help.

values = {'file': open(file_path, 'rb')}

poster.streaminghttp.register_openers()

datagen, headers = poster.encode.multipart_encode(values)

# Create the Request object
request = urllib2.Request(address, datagen, headers)
Liram Zarrouk
  • 9
  • 5

1 Answers1

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I have managed to solve this issue using the following code:

    f = open(file_path, "rb")
    mp = MultipartParam("file", fileobj=f, filename="file.csv")

    values = {"file": mp,
              constants.AUTHORIZATION_HEADER_KEY: constants.AUTHORIZATION_HEADER_VALUE % self._token,
              constants.CACHE_CONTROL_HEADER_KEY: constants.CACHE_CONTROL_NEGATIVE_VALUE
              }

    poster.streaminghttp.register_openers()

    datagen, headers = poster.encode.multipart_encode(values)

    # Create the Request object
    request = urllib2.Request(address, datagen, headers)
Liram Zarrouk
  • 9
  • 5