python shutil.move getting .filepart on my filename

Question

im upload a file to api with the following code

           files=[
              ('nzb_file',(nzbTitle,open(event.src_path,'rb'),'application/octet-stream'))
            ]
            headers = {}

            response = requests.request("POST", url, headers=headers, data=payload, files=files)

if response = 200 I want to move the file to DIR succes
if response contains something else. I want the file moved to another DIR

my problems is, Then I use shutil.move(xUpFile, nzbUpload) i get this error
FileNotFoundError: [Errno 2] No such file or directory: '/home/henrik/old_bot/Nzb/xxxxxxx.nzb.filepart'
I get this at the end of my file
.filepart

why and how do i solve this

I don’t now if it’s because the file still is open

Asked By: PeterGlad

||

Source

Answer 1

Requests doesn’t close the file for you, so you’ll need to call close() yourself, which means saving a reference to the file object. The following should work in Python 3.8+

    files=[
              ('nzb_file',(nzbTitle,f:=open(event.src_path,'rb'),'application/octet-stream'))
            ]
    headers = {}

    response = requests.request("POST", url, headers=headers, data=payload, files=files)
    f.close()

In versions before 3.8, you need to save off the file object reference before creating files:

    f = open(event.src_path,'rb')
    files=[
              ('nzb_file',(nzbTitle,f,'application/octet-stream'))
            ]
    headers = {}

    response = requests.request("POST", url, headers=headers, data=payload, files=files)
    f.close()

In either version, using a with block could also work since you’re only sending a single file, and the context manager will close the file for you. However this will also automatically catch any exceptions that are thrown within the block so in some cases you might not want that option.

    with open(event.src_path,'rb') as f:
        files=[
              ('nzb_file',(nzbTitle,f,'application/octet-stream'))
              ]
        headers = {}

        response = requests.request("POST", url, headers=headers, data=payload, files=files)

Also, unless you were specifically uploading a file with the .filepart extension, that is a temporary file created as part of the FTP process which should automatically get renamed or deleted by the FTP process when the transfer is successfully completed, so the problem may be that you’re trying to move the temporary file which the system has already cleaned up.

Answered By: nigh_anxiety

python shutil.move getting .filepart on my filename

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