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Python doesn't release file after it is closed

What I need to do is to write some messages on a .txt file, close it and send it to a server. This happens in a infinite loop, so the code should look more or less like this:
from requests_toolbelt.multipart.encoder import MultipartEncoder
num = 0
while True:
num += 1
filename = f"example{num}.txt"
with open(filename, "w") as f:
f.write("Hello")
f.close()
mp_encoder = MultipartEncoder(
fields={
'file': ("file", open(filename, 'rb'), 'text/plain')
}
)
r = requests.post("my_url/save_file", data=mp_encoder, headers=my_headers)
time.sleep(10)
The post works if the file is created manually inside my working directory, but if I try to create it and write on it through code, I receive this response message:
500 - Internal Server Error
System.IO.IOException: Unexpected end of Stream, the content may have already been read by another component.
I don't see the file appearing in the project window of PyCharm...I even used time.sleep(10) because at first, I thought it could be a time-related problem, but I didn't solve the problem. In fact, the file appears in my working directory only when I stop the code, so it seems the file is held by the program even after I explicitly called f.close(): I know the with function should take care of closing files, but it didn't look like that so I tried to add a close() to understand if that was the problem (spoiler: it was not)

I solved the problem by using another file
with open(filename, "r") as firstfile, open("new.txt", "a+") as secondfile:
secondfile.write(firstfile.read())
with open(filename, 'w'):
pass
r = requests.post("my_url/save_file", data=mp_encoder, headers=my_headers)
if r.status_code == requests.codes.ok:
os.remove("new.txt")
else:
print("File not saved")
I make a copy of the file, empty the original file to save space and send the copy to the server (and then delete the copy). Looks like the problem was that the original file was held open by the Python logging module

Firstly, can you change open(f, 'rb') to open("example.txt", 'rb'). In open, you should be passing file name not a closed file pointer.
Also, you can use os.path.abspath to show the location to know where file is written.
import os
os.path.abspath('.')
Third point, when you are using with context manager to open a file, you don't close the file. The context manger supposed to do it.
with open("example.txt", "w") as f:
f.write("Hello")

Close already open csv in Python

Is there a way for Python to close that the file is already open file.
Or at the very least display a popup that file is open or a custom written error message popup for permission error.
As to avoid:
PermissionError: [Errno 13] Permission denied: 'C:\\zf.csv'
I've seen a lot of solutions that open a file then close it through python. But in my case. Lets say I left my csv open and then tried to run the job.
How can I make it so it closes the currently opened csv?
I've tried the below variations but none seem to work as they expect that I have already opened the csv at an earlier point through python. I suspect I'm over complicating this.
f = 'C:\\zf.csv'
file.close()
AttributeError: 'str' object has no attribute 'close'
This gives an error as there is no reference to opening of file but simply strings.
Or even..
theFile = open(f)
file_content = theFile.read()
# do whatever you need to do
theFile.close()
As well as:
fileobj=open('C:\\zf.csv',"wb+")
if not fileobj.closed:
print("file is already opened")
How do I close an already open csv?
The only workaround I can think of would be to add a messagebox, though I can't seem to get it to detect the file.
filename = "C:\\zf.csv"
if not os.access(filename, os.W_OK):
print("Write access not permitted on %s" % filename)
messagebox.showinfo("Title", "Close your CSV")

Try using a with context, which will manage the close (__exit__) operation smoothly at the end of the context:
with open(...) as theFile:
file_content = theFile.read()

You can also try to copy the file to a temporary file, and open/close/remove it at will. It requires that you have read access to the original, though.
In this example I have a file "test.txt" that is write-only (chmod 444) and it throws a "Permission denied" error if I try writing to it directly. I copy it to a temporary file that has "777" rights so that I can do what I want with it:
import tempfile, shutil, os
def create_temporary_copy(path):
temp_dir = tempfile.gettempdir()
temp_path = os.path.join(temp_dir, 'temp_file_name')
os.chmod(temp_path, 0o777); # give full access to the tempfile so we can copy
shutil.copy2(path, temp_path) # copy the original into the temp one
os.chmod(temp_path, 0o777); # replace permissions from the original file
return temp_path
path = "./test.txt" # original file
copy_path = create_temporary_copy(path) # temp copy
with open(copy_path, "w") as g: # can do what I want with it
g.write("TEST\n")

f = open("C:/Users/amol/Downloads/result.csv", "r")
print(f.readlines()) #just to check file is open
f.close()
# here you can add above print statement to check if file is closed or not. I am using python 3.5

Read file using urllib and write adding extra characters

I have a script that regularly reads a text file on a server and over writes a copy of the text to a local copy of the text file. I have an issue of the process adding extra carriage returns and an extra invisible character after the last character. How do I make an identical copy of the server file?
I use the following to read the file
for link in links:
try:
f = urllib.urlopen(link)
myfile = f.read()
except IOError:
pass
and to write it to the local file
f = open("C:\\localfile.txt", "w")
try:
f.write(myfile)
except NameError:
pass
finally:
f.close()
This is how the file looks on the server
!http://i.imgur.com/rAnUqmJ.jpg
and this is how the file looks locally. Besides, an additional invisible character after the last 75
!http://i.imgur.com/xfs3E8D.jpg
I have seen quite a few similar questions, but not sure how to handle the urllib to read in binary
Any solution please?

If you want to copy a remote file denoted by a URL to a local file i would use urllib.urlretrieve:
import urllib
urllib.urlretrieve("http://anysite.co/foo.gz", "foo.gz")

I think urllib is reading binary.
Try changing
f = open("C:\\localfile.txt", "w")
to
f = open("C:\\localfile.txt", "wb")

Serving binary file from web server to client

Usually, when I want to transfer a web server text file to client, here is what I did
import cgi
print "Content-Type: text/plain"
print "Content-Disposition: attachment; filename=TEST.txt"
print
filename = "C:\\TEST.TXT"
f = open(filename, 'r')
for line in f:
print line
Works very fine for ANSI file. However, say, I have a binary file a.exe (This file is in web server secret path, and user shall not have direct access to that directory path). I wish to use the similar method to transfer. How I can do so?
What content-type I should use?
Using print seems to have corrupted content received at client side. What is the correct method?
I use the following code.
#!c:/Python27/python.exe -u
import cgi
print "Content-Type: application/octet-stream"
print "Content-Disposition: attachment; filename=jstock.exe"
print
filename = "C:\\jstock.exe"
f = open(filename, 'rb')
for line in f:
print line
However, when I compare the downloaded file with original file, it seems there is an extra whitespace (or more) for after every single line.

Agree with the above posters about 'rb' and Content-Type headers.
Additionally:
for line in f:
print line
This might be a problem when encountering \n or \r\n bytes in the binary file. It might be better to do something like this:
import sys
while True:
data = f.read(4096)
sys.stdout.write(data)
if not data:
break
Assuming this is running on windows in a CGI environment, you will want to start the python process with the -u argument, this will ensure stdout isn't in text-mode

When opening a file, you can use open(filename, 'rb') - the 'b' flag marks it as binary. For a general handler, you could use some form of mime magic (I'm not familiar with using it from Python, I've only ever used it from PHP a couple of years ago). For the specific case, .exe is application/octet-stream.

Content-type of .exe is tipically application/octet-stream.
You might want to read your file using open(filename, 'rb') where b means binary.
To avoid the whitespace problem, you could try with:
sys.stdout.write(open(filename,"rb").read())
sys.stdout.flush()
or even better, depending on the size of your file, use the Knio approach:
fo = open(filename, "rb")
while True:
buffer = fo.read(4096)
if buffer:
sys.stdout.write(buffer)
else:
break
fo.close()

For anyone using Windows Server 2008 or 2012 and Python 3, here's an update...
After many hours of experimentation I have found the following to work reliably:
import io
with io.open(sys.stdout.fileno(),"wb") as fout:
with open(filename,"rb") as fin:
while True:
data = fin.read(4096)
fout.write(data)
if not data:
break

Python ftplib: Overwriting a File doesn't work with STOR

I want to overwrite an existing file "test.txt" on my ftp server with this code:
from ftplib import FTP
HOST = 'host.com'
FTP_NAME = 'username'
FTP_PASS = 'password'
ftp = FTP(HOST)
ftp.login(FTP_NAME, FTP_PASS)
file = open('test.txt', 'r')
ftp.storlines('STOR test.txt', file)
ftp.quit()
file.close()
I don't get any error messages and the file test.txt has NOT been overwritten (the old test.txt is still on the server). I thought STOR overwrites files... Can somebody please help?
Thanks!

nvm, it's my fault...
I forgot to change the current working directory to /public_html
thanks anyway!

I think you need to open the file in write mode
file = open('test.txt', 'w')