How to extract a zip file using python 3

Question

How to extract a zip file using python when zip file present in different directory where script file present. I try this ,but i got error because source path is not accepted ,try to solve me this problem.

from zipfile import ZipFile

def func(source, target):
    with ZipFile('source', 'target'):
        ZipFile.Extractall('target')

Possible duplicate of [Extracting .zip in python](https://stackoverflow.com/questions/46994589/extracting-zip-in-python) — sophros, Jul 09 '19 at 07:18

score 7 · Answer 1 · answered Mar 06 '22 at 14:50

7

For just unpacking, shutil should suffice:

import shutil
shutil.unpack_archive('path-to-zipfile')

answered Mar 06 '22 at 14:50

00schneider

698
9
21

Best answer in this thread. – Arka Mukherjee May 20 '22 at 19:21

score 5 · Answer 2 · answered Jul 09 '19 at 07:20

Use this code. To move through directories you could either hard code the directory where your script is present or you could toggle through the directories using simple commands such as "../" to move out of the given directory or "/" to move inside a folder in the directory. For example - "../script.py" or "/folder/script.py". Similarly you can use this to find your .zip file.

import zipfile
with zipfile.ZipFile("file.zip","r") as zip_ref:
    zip_ref.extractall("targetdir")

score 1 · Answer 3 · answered Jul 09 '19 at 07:14

1

You'll have to check for source path of the zip file which is relative to your current working directory. To know your current working directory you can try

import os print(os.getcwd())

zip - Unzipping files in python

relative-paths-in-python

answered Jul 09 '19 at 07:14

Naman Chikara

164
14

How to extract a zip file using python 3

3 Answers3