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How to unzip many files with python

I got more than 1000 zip files in the same folder with naming convention output_MOJIBAKE
Example name: output_0aa3199eca63522b520ecfe11a4336eb_20210122_181742
How can I unzip them using Python?

Try this and let me know if it worked.
import os
import zipfile
path = 'path/to/your/zip/files'
os.chdir(path)
for file in os.listdir('.'):
with zipfile.ZipFile(file, 'r') as zip_ref:
zip_ref.extractall('.')

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.

Zip single file

I am trying to zip a single file in python. For whatever reason, I'm having a hard time getting down the syntax. What I am trying to do is keep the original file and create a new zipped file of the original (like what a Mac or Windows would do if you archive a file).
Here is what I have so far:
import zipfile
myfilepath = '/tmp/%s' % self.file_name
myzippath = myfilepath.replace('.xml', '.zip')
zipfile.ZipFile(myzippath, 'w').write(open(myfilepath).read()) # does not zip the file properly

The correct way to zip file is:
zipfile.ZipFile('hello.zip', mode='w').write("hello.csv")
# assume your xxx.py under the same dir with hello.csv
The python official doc says:
ZipFile.write(filename, arcname=None, compress_type=None)
Write the file named filename to the archive, giving it the archive name arcname
You pass open(filename).read() into write(). open(filename).read() is a single string that contains the whole content of file filename, it would throw FileNotFoundError because it is trying to find a file named with the string content.

If the file to be zipped (filename) is in a different directory called pathname, you should use the arcname parameter. Otherwise, it will recreate the full folder hierarchy to the file folder.
from zipfile import ZipFile
import os
with ZipFile(zip_file, 'w') as zipf:
zipf.write(os.path.join(pathname,filename), arcname=filename)

Try calling zipfile.close() afterwards?
from zipfile import ZipFile
zipf = ZipFile("main.zip","w", zipfile.ZIP_DEFLATED)
zipf.write("main.json")
zipf.close()

Since you also want to specify the directory try using os.chdir:
#!/usr/bin/python
from zipfile import ZipFile
import os
os.chdir('/path/of/target/and/destination')
ZipFile('archive.zip', 'w').write('original_file.txt')
Python zipfile : Work with Zip archives
Python Miscellaneous operating system interfaces

How to extract only mp3 files from a ZIP archive

I have this code:
from zipfile import ZipFile
import os
import glob
inp = raw_input("Specify a ZIP archive to extract:")
with ZipFile(inp) as zf:
zf.extractall()
It works fine because it extracts all the files but how do I extract all the .mp3 files in the archive that the user specifies.

To extract just the MP3 files from a ZIP archive, you could do the following:
from zipfile import ZipFile
import os
zip_file = r"c:\folder\myzip.zip"
target_folder = r"C:\Users\Fred\Desktop"
with ZipFile(zip_file, 'r') as my_zip:
mp3_files = [name for name in my_zip.namelist() if os.path.splitext(name)[1].lower() == '.mp3']
my_zip.extractall(target_folder, mp3_files)
The list of files inside the ZIP file can be obtained using the namelist function. With this you can filter just those files ending with an mp3 extension. The extractall function lets you pass a list of all of the files you want to extract (it defaults to all files).

You could get a list of the names of the members in the list, and only extract those ending with the suffix .mp3.