CODEWITHSUNDEEP
pythonunzip
newGIS
newGIS

Reputation: 614

Unzip zip files in folders and subfolders

I try to unzip 150 zip files. All the zip files as different names, and they all spread in one big folder that divided to a lot of sub folders and sub sub folders.i want to extract each archive to separate folder with the same name as the original zip file name and also in the same place as the original zip file . my code is:

import zipfile    
import os,os.path,sys  

pattern = '*.zip'  
folder = r"C:\Project\layers"   
files_process = []  
for root,dirs,files in os.walk(r"C:\Project\layers"):  
    for filenames in files:  
        if filenames == pattern:  
            files_process.append(os.path.join(root, filenames))  
            zip.extract()

After i run the code nothing happened. Thanks in advance for any help on this.

Upvotes: 7

Views: 37744

Answers (3)

George Fisher
George Fisher

Reputation: 3344

To unzip all the files into a temporary folder (Ubuntu)

import tempfile
import zipfile

tmpdirname = tempfile.mkdtemp()

zf = zipfile.ZipFile('/path/to/zipfile.zip')

for fn in zf.namelist():
    temp_file = tmpdirname+"/"+fn
    #print(temp_file)

    f = open(temp_file, 'w')
    f.write(zf.read(fn).decode('utf-8'))
    f.close()

Upvotes: -1

newGIS
newGIS

Reputation: 614

UPDATE:

Finally, this code worked for me:

import zipfile,fnmatch,os

rootPath = r"C:\Project"
pattern = '*.zip'
for root, dirs, files in os.walk(rootPath):
    for filename in fnmatch.filter(files, pattern):
        print(os.path.join(root, filename))
        zipfile.ZipFile(os.path.join(root, filename)).extractall(os.path.join(root, os.path.splitext(filename)[0]))

Upvotes: 16

jfs
jfs

Reputation: 414205

You could use Path.rglob() to enumerate zip-files recursively and shutil.unpack_archive() to unpack zip files:

#!/usr/bin/env python3
import logging
from pathlib import Path
from shutil import unpack_archive

zip_files = Path(r"C:\Project\layers").rglob("*.zip")
while True:
    try:
        path = next(zip_files)
    except StopIteration:
        break # no more files
    except PermissionError:
        logging.exception("permission error")
    else:
         extract_dir = path.with_name(path.stem)
         unpack_archive(str(path), str(extract_dir), 'zip')

It "extract[s] each archive to separate folder with the same name as the original zip file name and also in the same place as the original zip file" e.g., it extracts 'layers/dir/file.zip' archive into 'layers/dir/file' directory.

Upvotes: 6

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