Use subprocess python library to unzip a file using 7zip

Question

I would like to unzip a file with Python using 7zip executable. In Perl this is pretty straightforward:

$zip_exe_path = "C:\\Dropbox\\7-zip\\7z.exe";
$logfile_path = "C:\\Temp\\zipped_file.7z";
system ("$zip_exe_path x $log_file_path -y");

I tried this:

import subprocess
zip_exe_path = "C:\\Dropbox\\7-zip\\7z.exe"
logfile_path = "C:\\Temp\\zipped_file.7z"
subprocess.call(['zip_exe_path','x','logfile_path','-y'])

When I do so I get this error:

FileNotFoundError: [WinError 2] The system cannot find the file specified

Thanks for any help!

midori · Answer 1 · 2016-03-02T18:28:18.190

3

Why not use python zip:

import zipfile

with zipfile.ZipFile(logfile_path, 'r') as z:
    z.extractall()

Or using subprocess:

subprocess.call(['zip_exe_path','x','logfile_path','-y'], shell=True)

edited Mar 02 '16 at 18:28

answered Mar 02 '16 at 18:11

midori

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Thanks for the advice. That doesn't work because zipfile doesn't work on other .zip formats like .7z or .rar. – sparrow Mar 02 '16 at 18:18
Use py7zlib in such case – midori Mar 02 '16 at 18:20
I did but it has problems handling corrupt files. Which is why I would like to use system commands. – sparrow Mar 02 '16 at 18:22
in that case add shell=True to your subprocess call – midori Mar 02 '16 at 18:26
if downvote please leave a comment, so i would know how to correct the answer, otherwise your downvote is useless – midori Mar 03 '16 at 23:45
we can specify the specific path in `z.extractall(path)` to extract the files at a particular location. – mnis.p Jul 06 '18 at 10:47

score 3 · Answer 2 · answered Oct 29 '20 at 08:09

This work like a charm for me :)

first, install py7zr library:

pip install py7zr

for extracting all files in .7z:

from py7zr import py7zr

with py7zr.SevenZipFile('7z file_location', mode='r') as z:
    z.extractall()

for extracting a single file:

from py7zr import py7zr

with py7zr.SevenZipFile('7z file_location', mode='r') as z:
    z.extract(targets=['rootdir/filename'])

score 2 · Accepted Answer · answered Jan 02 '20 at 09:57

The error is that you are passing the strings 'zip_exe_path' and 'logfile_path' instead of the values of those variables.

import subprocess
zip_exe_path = "C:\\Dropbox\\7-zip\\7z.exe"
logfile_path = "C:\\Temp\\zipped_file.7z"
subprocess.call([zip_exe_path, 'x', logfile_path, '-y'])

You can of course pass the command as a single string with shell=True but the shell does not add any value and incurs some overhead (and risk!)

score 1 · Answer 4 · answered Mar 03 '16 at 02:03

1

Figured it out:

import subprocess

subprocess.Popen(zip_exe+' x '+file+' -o'+output_loc,stdout=subprocess.PIPE)

answered Mar 03 '16 at 02:03

sparrow

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Switching from `call` to `Popen` is not an improvement here. Requiring a shell is also not an improvement. – tripleee Jan 02 '20 at 09:59

Use subprocess python library to unzip a file using 7zip

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