Permission error when unzipping files that are password protected

Question

I am trying to unzip a number of files that are password protected but I keep getting some permission error. I have tried to perform this operation running vscode as an administrator but I am still getting the same error.

Here is the code:

input_file = ".\\pa-dirty-price-crawler\\folders"

import zipfile
with zipfile.ZipFile(input_file, 'r') as zip_ref:
    zip_ref.extractall(input_file, pwd=b'qpsqpwsr')

Here is the error:

Traceback (most recent call last):
  File "c:/Users/usr/workspace/pa-dirty-price-crawler/src/outlook.py", line 23, in <module>
    with zipfile.ZipFile(input_file, 'r') as zip_ref:
  File "C:\ProgramData\Anaconda3\lib\zipfile.py", line 1240, in __init__
    self.fp = io.open(file, filemode)
PermissionError: [Errno 13] Permission denied: '.\\pa-dirty-price-crawler\\folders'

I do not know of another library that can do this same operation but if anyone has suggestions in regards to getting this fixed I would really appreciate it.

Edit:

When I try to specify the entire file path name as so:

input_file = "C:\\Users\\usr\\workspace\\pa-dirty-price-crawler\\folders"

import zipfile
with zipfile.ZipFile(input_file, 'r') as zip_ref:
    zip_ref.extractall(pwd=b'qpsqpwsr')

I still get this error:

Traceback (most recent call last):
  File "c:/Users/usr/workspace/pa-dirty-price-crawler/src/outlook.py", line 23, in <module>
    with zipfile.ZipFile(input_file, 'r') as zip_ref:
  File "C:\ProgramData\Anaconda3\lib\zipfile.py", line 1240, in __init__
    self.fp = io.open(file, filemode)
PermissionError: [Errno 13] Permission denied: 'C:\\Users\\usr\\workspace\\pa-dirty-price-crawler\\folders'

Your example code looks like your input is a directory not a zip file. Then you're trying to extract to the same directory. Are you sure your example is correct? — Alastair McCormack, Apr 13 '20 at 16:13
@PatrickArtner Yes that string works when I put in that password manually. — Snorrlaxxx, Apr 13 '20 at 16:19
@AlastairMcCormack I am not sure if it is correct should I leave the path for the extract file none for now? — Snorrlaxxx, Apr 13 '20 at 16:19

Alastair McCormack · Accepted Answer · 2020-04-13T16:56:33.600

2

It looks like you're passing a directory as the input. This is the likely problem, not that the zip is password protected.

To extract a zip file, zipfile.ZipFile takes a zip file as an input not a directory.

Therefore, your code needs two variables: an input zip file and an output directory:

input_file = r".\pa-dirty-price-crawler\folders\myzipfile.zip"
output_directory = r".\pa-dirty-price-crawler\folders"

import zipfile
with zipfile.ZipFile(input_file, 'r') as zip_ref:
    zip_ref.extractall(output_directory, pwd=b'qpsqpwsr')

* note the use of r"string", this helps having to escape all your back slashes

edited Apr 13 '20 at 16:56

answered Apr 13 '20 at 16:50

Alastair McCormack

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That worked, than you very much. Although is it possible to do this for multiple zip files instead of specifying the \myzipfile.zip in the input_file path – Snorrlaxxx Apr 13 '20 at 17:51
I hope my question made sense there – Snorrlaxxx Apr 13 '20 at 17:54
1

Use `Path.glob("directory/*.zip")` with a `for` loop – Alastair McCormack Apr 13 '20 at 18:10

Permission error when unzipping files that are password protected

1 Answers1