CODEWITHSUNDEEP
pythonjarunzipzip
Ankur Agarwal
Ankur Agarwal

Reputation: 24758

Extracting .zip in python

I got BadZipfile: Bad magic number for file header error while extracting a .zip using python2 zipfile.ZipFile

Same .zip when extracted with unzip gives file #1: bad zipfile offset (local header sig): 0 but gets extracted with exit code 2.

When using jar -xf file.zip the command completes with $? == 0 with nothing being extracted.

Using file gives:

file -i file.zip
file.zip application/octet-stream; charset=binary

This gives incorrect header for zipfile

$ hexdump -C file.zip | head -10
00000000  50 67 f0 de 1e 7a 29 e4  93 56 3f 11 a2 5f b6 97  |Pg...z)..V?.._..|

Correct header is:

00000000  50 4b 03 04 14 00 08 08  08 00 28 3e 4b 4b 00 00  |PK........(>KK..|

Why is the file listed as application/octet-stream ?

I am on

Distributor ID: Ubuntu
Description:    Ubuntu 14.04.5 LTS
Release:    14.04
Codename:   trusty

Whats going on ? What file format is this ? Any pointers ?

Upvotes: 1

Views: 1635

Answers (2)

amit dorman
amit dorman

Reputation: 1

You can try this:

def open_zip(path_to_archive, folder_arg):
    if zipfile.is_zipfile(path_to_archive) is True:
        zip_ref = zipfile.ZipFile(path_to_archive, 'r')
        zip_ref.extractall(folder_arg)
        zip_ref.close()
        print("finish to extract zip file")
    else:
        print(f"the file {path_to_archive} is not a zip file")

Upvotes: 0

Josef B.
Josef B.

Reputation: 249

Have you tried this?

import zipfile
zip_ref = zipfile.ZipFile(path_to_zip_file, 'r')
zip_ref.extractall(directory_to_extract_to)
zip_ref.close()

Upvotes: 1

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