Develop Reference

tarfile.open() does not extract into the right directory path

I'm trying to extract all from a tar.gz file into the same Directory. The following code works to extract all, but the files are stored in the working directory instead of the path I entered as name.
import tarfile
zip_rw_data = r"P:\Lehmann\Test_Python_Project\RW_data.tar.gz"
tar = tarfile.open(name=zip_rw_data, mode='r')
tar.extractall()
tar.close()
How do I make sure the extracted files are saved in the directory path where I need them? I've been trying at this for ages, I really can't see why this doesn't work.

You should use:
import tarfile
zip_rw_data = r"P:\Lehmann\Test_Python_Project\RW_data.tar.gz"
tar = tarfile.open(name=zip_rw_data, mode='r')
tar.extractall(path=r"P:\Lehmann\Test_Python_Project")
tar.close()

You can try using shutil.unpack_archive
def extract_all(archives, extract_path):
for filename in archives:
shutil.unpack_archive(filename, extract_path)

How can I extract all .zip extension in a folder without retaining directory using python?

Here is my code I don't know how can I loop every .zip in a folder, please help me: I want all contents of 5 zip files to extracted in one folder, not including its directory name
import os
import shutil
import zipfile
my_dir = r"C:\\Users\\Guest\\Desktop\\OJT\\scanner\\samples_raw"
my_zip = r"C:\\Users\\Guest\\Desktop\\OJT\\samples\\001-100.zip"
with zipfile.ZipFile(my_zip) as zip_file:
zip_file.setpassword(b"virus")
for member in zip_file.namelist():
filename = os.path.basename(member)
# skip directories
if not filename:
continue
# copy file (taken from zipfile's extract)
source = zip_file.open(member)
target = file(os.path.join(my_dir, filename), "wb")
with source, target:
shutil.copyfileobj(source, target)

repeated question, please refer below link.
How to extract zip file recursively in Pythonn

What you are looking for is glob. Which can be used like this:
#<snip>
import glob
#assuming all your zip files are in the directory below.
for my_zip in glob.glob(r"C:\\Users\\Guest\\Desktop\\OJT\\samples\\*.zip"):
with zipfile.ZipFile(my_zip) as zip_file:
zip_file.setpassword(b"virus")
for member in zip_file.namelist():
#<snip> rest of your code here.

Python:writing files in a folder to zipfile and compressing it

I am trying to zip and compress all the files in a folder using python. The eventual goal is to make this occur in windows task scheduler.
import os
import zipfile
src = ("C:\Users\Blah\Desktop\Test")
os.chdir=(src)
path = (r"C:\Users\Blah\Desktop\Test")
dirs = os.listdir(path)
zf = zipfile.ZipFile("myzipfile.zip", "w", zipfile.ZIP_DEFLATED,allowZip64=True)
for file in dirs:
zf.write(file)
Now when I run this script I get the error:
WindowsError: [Error 2] The system cannot find the file specified: 'test1.bak'
I know it's there since it found the name of the file it can't find.
I'm wondering why it is not zipping and why this error is occurring.
There are large .bak files so they could run above 4GB, which is why I'm allowing 64 bit.
Edit: Success thanks everyone for answering my question here is my final code that works, hope this helps my future googlers:
import os
import zipfile
path = (r"C:\Users\vikram.medhekar\Desktop\Launch")
os.chdir(path)
dirs = os.listdir(path)
zf = zipfile.ZipFile("myzipfile.zip", "w", zipfile.ZIP_DEFLATED,allowZip64=True)
for file in dirs:
zf.write(os.path.join(file))
zf.close()

os.listdir(path) returns names relative to path - you need to use zf.write(os.path.join(path, file)) to tell it the full location of the file.

As I said twice in my comments:
Python is not looking for the file in the folder that it's in, but in the current working directory. Instead of
zf.write(file)
you'll need to
zf.write(path + os.pathsep + file)

How to compress a file with shutil.make_archive in python?

I want to compress one text file using shutil.make_archive command. I am using the following command:
shutil.make_archive('gzipped'+fname, 'gztar', os.path.join(os.getcwd(), fname))
OSError: [Errno 20] Not a directory: '/home/user/file.txt'
I tried several variants but it keeps trying to compress the whole folders. How to do it correctly?

Actually shutil.make_archive can make one-file archive! Just pass path to target directory as root_dir and target filename as base_dir.
Try this:
import shutil
file_to_zip = 'test.txt' # file to zip
target_path = 'C:\\test_yard\\' # dir, where file is
try:
shutil.make_archive(target_path + 'archive', 'zip', target_path, file_to_zip)
except OSError:
pass

shutil can't create an archive from one file. You can use tarfile, instead:
tar = tarfile.open(fname + ".tar.gz", 'w:qz')
os.chdir('/home/user')
tar.add("file.txt")
tar.close()
or
tar = tarfile.open(fname + ".tar.gz", 'w:qz')
tar.addfile(tarfile.TarInfo("/home/user/file.txt"), "/home/user/file.txt")
tar.close()

Try this and Check shutil
copy your file to a directory.
cd directory
shutil.make_archive('gzipped', 'gztar', os.getcwd())

#CommonSense had a good answer, but the file will always be created zipped inside its parent directories. If you need to create a zipfile without the extra directories, just use the zipfile module directly
import os, zipfile
inpath = "test.txt"
outpath = "test.zip"
with zipfile.ZipFile(outpath, "w", compression=zipfile.ZIP_DEFLATED) as zf:
zf.write(inpath, os.path.basename(inpath))

If you don't mind doing a file copy op:
def single_file_to_archive(full_path, archive_name_no_ext):
tmp_dir = tempfile.mkdtemp()
shutil.copy2(full_path, tmp_dir)
shutil.make_archive(archive_name_no_ext, "zip", tmp_dir, '.')
shutil.rmtree(tmp_dir)

Archiving a directory to another destination was a pickle for me but shutil.make_archive not zipping to correct destination helped a lot.
from shutil import make_archive
make_archive(
base_name=path_to_directory_to_archive},
format="gztar",
root_dir=destination_path,
base_dir=destination_path)

How to unzip specific folder from a .zip with Python

I am looking to unzip a particular folder from a .zip in Python:
e.g. archive.zip contains the folders foo and bar, I want to unzip foo to a specific location, retaining it's folder structure.

Check zipfile module.
For your case:
import zipfile
archive = zipfile.ZipFile('archive.zip')
for file in archive.namelist():
if file.startswith('foo/'):
archive.extract(file, 'destination_path')

You should close your zips....
import zipfile
archive = zipfile.ZipFile('archive.zip')
for file in archive.namelist():
if file.startswith('foo/'):
archive.extract(file, 'destination_path')
archive.close()
Or just use a safer method. With will close your zip.
import zipfile
with zipfile.ZipFile('archive.zip') as archive:
for file in archive.namelist():
if file.startswith('foo/'):
archive.extract(file, 'destination_path')

I like to reduce the list of names first so that the for loop doesn't parse through all the files in the zip archive:
import zipfile
archive = zipfile.ZipFile('archive.zip')
names_foo = [i for i in archive.namelist() if i.startswith('foo') ]
for file in names_foo:
archive.extract(file)

using zipfile library is very very slow.
this is better way:
os.system('unzip -P your-password path/to/file.zip')