Python 3.6.1 - PermissionError: [Errno 13] Permission denied shown when trying to unzip a file

Question

I am facing a PermissionError while I am trying to extract a zip file. I have gone through a lot of discussion threads here on SO but still unable to solve my issue.

Currently I am working with Python 3.6.1 on a Windows 8 box. I have created a new directory through the following code:

import os,zipfile

newpath = 'C:\\home\\vivvin\\shKLSE'
#newpath = r'C:\\home\\vivvin\\shKLSE'
if not os.path.exists(newpath):
os.makedirs(newpath)

Next I have downloaded a zip file and saved into newpath directory.

Now I am trying to extract all the files (10 csv files) within the zip file to be extracted into the newpath directory. To achieve that I have written the following code:

import os,zipfile

newpath = 'C:\\home\\vivvin\\shKLSE'
path_to_zip_file = newpath
directory_to_extract_to = newpath
#zip_ref = zipfile.ZipFile(newpath, 'r')
zip_ref = zipfile.ZipFile(newpath, 'w')
zip_ref.extractall(newpath)
zip_ref.close()

But each time I am getting an error as:

Traceback (most recent call last):
File "C:/Users/AtechM_03/PycharmProjects/Webinar/SeleniumScripts/extract.py", line 6, in <module>
zip_ref = zipfile.ZipFile(newpath, 'w')
File "C:\Python\lib\zipfile.py", line 1082, in __init__
self.fp = io.open(file, filemode)
PermissionError: [Errno 13] Permission denied: 'C:\\home\\vivvin\\shKLSE'

I have observed the properties manually of the zip file and seems there is a Security message along with a Unblock button. As of now I am clueless how to Unblock it.

Can anyone help me out please? Thanks in advance.

Solution 1

I had a similar problem trying to write to a file.

The fix that worked for me:

Right-click your PyCharm application and run it as administrator.

Solution 2

I run into the same error when unzipping the folder "temp.zip" and only extract the file . In my case I had a directory with a folder "temp" and a zip file called "temp.zip".

def unzip(path, filename): 
    with ZipFile(path, 'r') as zipobj: 
        zipobj.extract(member=filename)

When I run this file, I got the error message:

    ...
   "test.py", line 259, in unzip
        with ZipFile(path, 'r') as zipobj: 
      File "C:\Program Files\Python36\lib\zipfile.py", line 1090, in __init__
        self.fp = io.open(file, filemode)
    PermissionError: [Errno 13] Permission denied

The problem is, that zipobj.extract() needs to create a folder called temp and extract the content (temporarly). But this folder already exists. --> There I got the permission denied error.

Solution:

• either delete the folder "temp"
• or move the zip first to another directory and unzip it there

Solution 3

I guess it's because your dictionary is on the C drive (the windows disk ,sometimes it's forbidden to write and erase) , if you changed to the D drive, it maybe work .

Comments

[Exprator]

Pycharm dont have permission, try opening cmd after right click and then run as administrator then go to the python program folder and run your file

[undetected Selenium]

@Exprator Thanks for the quick look. But I am doing a lot other stuff through PyCharm with Selenium. Is there a way I can achieve the permission? Thanks

[Exprator]

Well you can give permission while creating the folder. Check for os permission giving in python

[undetected Selenium]

@Exprator The permission issue is not with the directory but with the downloaded zip file. Thanks

[stovfl]

Where is the filename of the zip file in your script?

[undetected Selenium]

@stovfl Thanks a lot for pointing that out. What if I am not sure about the exact file name? But definitely I am sure about the zip type of file.

[stovfl]

Read about using *.zip: docs.python.org/3/library/glob.html#module-glob

[undetected Selenium]

@stovfl Thanks for pointing to the required document. I will get back to you after going through that. Thanks