Suposse that I have some files that I downloaded from a server and they are zipped with 7zip in multiple parts, the format is something like this myfile.zip.001, myfile.zip.002, ..., myfile.zip.00n. Basically, I need to extract the content of it in the same folder where they are stored.
I tried using zipfile, patoolib and pyunpack without success, here is what I've done:
file_path = r"C:\Users\user\Documents\myfile.zip.001" #I also tested with only .zip
extract_path = r"C:\Users\user\Documents\"
#"
import zipfile
with zipfile.ZipFile(file_path, "r") as zip_ref:
zip_ref.extractall(extract_path) # myfile.zip.001 file isn't zip file.
from pyunpack import Archive
Archive(file_path).extractall(extract_path) # File is not a zip file
import patoolib
patoolib.extract_archive(file_path, outdir=extract_path) # unknown archive format for file `myfile.zip.001'
Another way (that works, but it's very ugly) is this one:
import os
import subprocess
path_7zip = r"C:\Program Files (x86)\7-Zip\7z.exe"
cmd = [path_7zip, 'x', 'myfile.zip.001']
sp = subprocess.Popen(cmd, stderr=subprocess.STDOUT, stdout=subprocess.PIPE)
But this makes the user install 7zip in his computer, which isn't a good approach of what I'm looking for.
So, the question is: there is at least a way to extract/unzip multi-parts files with the format x.zip.001 in python?
You seem to be on the right track with zipfile, but you most likely have to concatenate the zip file before using extractall.
import os
zip_prefix = "myfile.zip."
# N number of parts
import glob
parts = glob.glob(zip_prefix + '*')
n = len(parts)
# Concatenate
with open("myfile.zip", "wb") as outfile:
for i in range(1, n+1):
filename = zip_prefix + str(i).zfill(3)
with open(filename, "rb") as infile:
outfile.write(infile.read())
# Extract
import zipfile
with zipfile.ZipFile(file_path, "r") as zip_ref:
zip_ref.extractall(extract_path)