Extracting .zip in python

Question

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 ?

Freefloating

have you tried all other ways: gzip, unzip ? – Ammonate

maybe archive is corrupted? – Managerial

Possible duplicate of Bad magic number error with ZipFile module in Python – Zeph

@Ammonate With unzip I get exit code 2 but the file gets extracted. – Freefloating

file -i file.zip gives application/octet-stream. Why ? – Freefloating

I added few extra details from hexdump. Which file format is this ? – Freefloating

Answer 1

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()

Porky

Then may be this could help [programcreek.com/python/example/3149/zipfile.BadZipfile] – Porky