How to unzip with 7z in Python

Question

I want to unzip a archive file in a special directory. i use subprocess as below :

import subprocess
INPUT_FILE = "../DEV-TESTSTATION/tmp/DLL_LOGFILE-G1R0C0.zip"
OUTPUT = "../DEV-TESTSTATION/work/LOGFILE"

subprocess.call(['7z', 'x', INPUT_FILE ,' -o'+OUTPUT ])

Output :

No files to process
Everything is Ok

Files: 0
Size:       0
Compressed: 2013846

why not just use [`zipfile`](https://docs.python.org/3/library/zipfile.html)? — Tomerikoo, Sep 01 '20 at 09:20
@ewong i get this error when i test your line : Too short switch: -o — , Sep 01 '20 at 09:23
note that there is [py7zr](https://github.com/miurahr/py7zr) — FObersteiner, Sep 01 '20 at 09:24

score 0 · Answer 1 · answered Sep 01 '20 at 09:33

0

You can try py7xr library.

Example:

import py7zr 

with py7zr.SevenZipFile('abc.7z', mode='r', password='secret') as 7z: 

  7z.extractall()

answered Sep 01 '20 at 09:33

Jimmy

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How to unzip with 7z in Python

1 Answers1