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Heroku cant download file using python

I need the bot to download the file from the link. I am using this function:
def download(url, filename):
get_response = requests.get(url,stream=True)
file_name = filename
with open(file_name, 'wb') as f:
for chunk in get_response.iter_content(chunk_size=1024):
if chunk: # filter out keep-alive new chunks
f.write(chunk)
And then I send it to the user using this:
with open(music, 'rb') as music_file:
await bot.send_audio(message.chat.id, music_file)
Everything works great on my computer. But when I run the bot on Heroku, it constantly gives this error:
aiogram.utils.exceptions.BadRequest: File must be non-empty
I tried to add delays, tried to check if the file was downloaded, but nothing helped. Can anyone help?

PermissionError: [WinError 5] Access is denied: 'C:\\Users\\testuser'

I'm trying to download a file from this website with python.
I however get this error:PermissionError: [WinError 5] Access is denied: 'C:\\Users\\testuser'
Note that I cannot run this code as admin. It has to be solved somehow programmatically
This is the code:
import os
import stat
import requests
def download(url_string: str, destination_folder: str):
if not os.path.exists(destination_folder):
os.makedirs(destination_folder) # create folder if it does not exist
filename = url_string.split('/')[-1].replace(" ", "_") # be careful with file names
file_path = os.path.join(destination_folder, filename)
r = requests.get(url_string, stream=True)
if r.ok:
print("saving to", os.path.abspath(file_path))
with open(file_path, 'wb') as f:
for chunk in r.iter_content(chunk_size=1024 * 8):
if chunk:
f.write(chunk)
f.flush()
os.fsync(f.fileno())
else: # HTTP status code 4XX/5XX
print("Download failed: status code {}\n{}".format(r.status_code, r.text))
url = r'https://www.dundeecity.gov.uk/sites/default/files/publications/civic_renewal_forms.zip'
path = r'C:\Users\testuser\Desktop\report\report.zip'
download(url, path)

Go to Start > Settings > Update & Security > Windows Security > Virus & threat protection.
then, click exclusions and add your python file.

I solved it!
First of all, I've apparently misspelled the username of the computer: it's test_user. Silly, I know, but difficult until you see it.
Second of all python apparently seems to want to force me to make this file a txt. Had to force it myself to make it a Zip.
Okay.
Glad it was fixed quickly. Thanks for the support :D

Azure - Files in file share with given extension

I'm using this to connect to Azure File Share and upload a file. I would like to chose what extension file will have, but I can't. I got an error shown below. If I remove .txt everything works fine. Is there a way to specify file extension while uploading it?
Error:
Exception: ResourceNotFoundError: The specified parent path does not exist.
Code:
def main(blobin: func.InputStream):
file_client = ShareFileClient.from_connection_string(conn_str="<con_string>",
share_name="data-storage",
file_path="outgoing/file.txt")
f = open('/home/temp.txt', 'w+')
f.write(blobin.read().decode('utf-8'))
f.close()
# Operation on file here
f = open('/home/temp.txt', 'rb')
string_to_upload = f.read()
f.close()
file_client.upload_file(string_to_upload)

I believe the reason you're getting this error is because outgoing folder doesn't exist in your file service share. I took your code and ran it with and without extension and in both situation I got the same error.
Then I created a folder and tried to upload the file and I was able to successfully do so.
Here's the final code I used:
from azure.storage.fileshare import ShareFileClient, ShareDirectoryClient
conn_string = "DefaultEndpointsProtocol=https;AccountName=myaccountname;AccountKey=myaccountkey;EndpointSuffix=core.windows.net"
share_directory_client = ShareDirectoryClient.from_connection_string(conn_str=conn_string,
share_name="data-storage",
directory_path="outgoing")
file_client = ShareFileClient.from_connection_string(conn_str=conn_string,
share_name="data-storage",
file_path="outgoing/file.txt")
# Create folder first.
# This operation will fail if the directory already exists.
print "creating directory..."
share_directory_client.create_directory()
print "directory created successfully..."
# Operation on file here
f = open('D:\\temp\\test.txt', 'rb')
string_to_upload = f.read()
f.close()
#Upload file
print "uploading file..."
file_client.upload_file(string_to_upload)
print "file uploaded successfully..."

Python file delete PermissionError: [WinError 32]

I am trying to delete a duplicated image by comparing md5 file hash.
my code is
from PIL import Image
import hashlib
import os
import sys
import io
img_file = urllib.request.urlopen(img_url, timeout=30)
f = open('C:\\Users\\user\\Documents\\ + img_name, 'wb')
f.write(img_file.read())
f.close # subject image, status = ok
im = Image.open('C:\\Users\\user\\Documents\\ + img_name)
m = hashlib.md5() # get hash
with io.BytesIO() as memf:
im.save(memf, 'PNG')
data = memf.getvalue()
m.update(data)
md5hash = m.hexdigest() # hash done, status = ok
im.close()
if md5hash in hash_list[name]: # comparing hash
os.remove('C:\\Users\\user\\Documents\\ + img_name) # delete file, ERROR
else:
hash_list[name].append(m.hexdigest())
and i get this error
PermissionError: [WinError 32] The process cannot access the file because it is being used by another process:
'C:\\Users\\user\\Documents\\myimage.jpg'
I tried admin command prompt, but still getting this error. Could you find what is accessing the file?

Just noticed you're using f.close instead of f.close()
Add () and check if problem still occurs.
Cheers ;)

Your issue has indeed been as Adrian Daniszewski said, however, there are quite few more programming problems with your code.
First of all, you should familiarize yourself with with. You use with for BytesIO() but it can also be used for opening files.
The benefit of with open(...) as f: is the fact that you don't have to search whether you closed the file or remember to close it. It will close the file at the end of its indentation.
Second, there is a bit of duplication in your code. Your code should be DRY to avoid being forced to change multiple locations with the same stuff.
Imagine having to change where you save the byte files. Right now you will be forced to change in three different locations.
Now imagine not noticing one of these locations.
My suggestion would be first of all to save the path to a variable and use that -
bytesimgfile = 'C:\\Users\\user\\Documents\\' + img_name
An example to use with in your code would be like this:
with open(bytesimgfile , 'wb') as f:
f.write(img_file.read())
A full example with your given code:
from PIL import Image
import hashlib
import os
import sys
import io
img_file = urllib.request.urlopen(img_url, timeout=30)
bytesimgfile = 'C:\\Users\\user\\Documents\\' + img_name
with open(bytesimgfile , 'wb'):
f.write(img_file.read())
with Image.open(bytesimgfile) as im:
m = hashlib.md5() # get hash
with io.BytesIO() as memf:
im.save(memf, 'PNG')
data = memf.getvalue()
m.update(data)
md5hash = m.hexdigest() # hash done, status = ok
if md5hash in hash_list[name]: # comparing hash
os.remove(bytesimgfile) # delete file, ERROR
else:
hash_list[name].append(m.hexdigest())

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