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Extracting zip with password to another dir without foldername

I have this password protected zip folder:
folder_1\1.zip
When I extract this it gives me
1\image.png
How can I extract this to another folder without its folder name? Just the contents of it: image.png
So far I have done all stackoverflows solutions and took me 11 hrs straight just to solve this.
import zipfile
zip = zipfile.ZipFile('C:\\Users\\Desktop\\folder_1\\1.zip', 'r')
zip.setpassword(b"virus")
zip.extractall('C:\\Users\\Desktop') <--target dir to extract all contents
zip.close()
EDIT:
This code worked for me: (Now I want many paths to be extracted at once, any ideas?
import os
import shutil
import zipfile
my_dir = r"C:\\Users\\Desktop"
my_zip = r"C:\\Users\\Desktop\\test\\folder_1\\1.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)

You can use the ZipFile.read() method to read the specific file in the archive, open your target file for writing by joining the target directory with the base name of the source file, and then write what you read to it:
import zipfile
import os
zip = zipfile.ZipFile('C:\\Users\\Desktop\\folder_1\\1.zip', 'r')
zip.setpassword(b"virus")
for name in zip.namelist():
if not name.endswith(('/', '\\')):
with open(os.path.join('C:\\Users\\Desktop', os.path.basename(name)), 'wb') as f:
f.write(zip.read(name))
zip.close()
And if you have several paths containing 1.zip for extraction:
import zipfile
import os
for path in 'C:\\Users\\Desktop\\folder_1', 'C:\\Users\\Desktop\\folder_2', 'C:\\Users\\Desktop\\folder_3':
zip = zipfile.ZipFile(os.path.join(path, '1.zip'), 'r')
zip.setpassword(b"virus")
for name in zip.namelist():
if not name.endswith(('/', '\\')):
with open(os.path.join('C:\\Users\\Desktop', os.path.basename(name)), 'wb') as f:
f.write(zip.read(name))
zip.close()

no such file of directory 'file.xml'

Having a small issues when trying to create a zip file using the zipfile module in python 3.
I have a directory which contains xml files, I am looking to create a zip archive from all these files in the same directory but keep encountering the error of FileNotFoundError: [Errno 2] no such file or directory: 'file.xml'
script:
import datetime
import os
import zipfile
path = '/Users/xxxx/reports/xxxx/monthly'
month = datetime.datetime.now().strftime('%G'+'-'+'%B')
zf = os.path.join(path, '{}.zip'.format(month))
z = zipfile.ZipFile(zf, 'w')
for i in os.listdir(path):
if i.endswith('.xml'):
z.write(i)
z.close()
it seems like when z.write(i) is called it is looking in the working directory for the xml files however the working directory is /Users/xxxx/scripts where the python script is.
How would I get the z.write(i) to look at the path variable without changing the current working directory if possible.

What actually happens is that as you loop through os.listdir(path), the i itself is simply the FileName which does not include the real Path to the File. There are a couple of ways to get around this; the simplest (but crudest) of which is shown below:
import datetime
import os
import zipfile
path = '/Users/xxxx/reports/xxxx/monthly'
month = datetime.datetime.now().strftime('%G'+'-'+'%B')
zf = os.path.join(path, '{}.zip'.format(month))
z = zipfile.ZipFile(zf, 'w')
for i in os.listdir(path):
# DECLARE A VARIABLE TO HOLD THE FULL PATH TO THE FILE:
xmlFile = "{}/{}".format(path, i) # <== PATH TO CURRENT FILE UNDER CURSOR
if xmlFile.endswith('.xml'):
z.write(xmlFile)
z.write(filename=xmlFile, arcname="ARCHIVE_NAME_HERE", ) # <== CHANGE
z.close()
Hope this Helps.
Cheers and Good-Luck...

use os.chdir to move to the file path and try to write the files to zip.
import datetime
import os
import zipfile
path = '/Users/xxxx/reports/xxxx/monthly'
month = datetime.datetime.now().strftime('%G'+'-'+'%B')
zf = os.path.join(path, '{}.zip'.format(month))
z = zipfile.ZipFile(zf, 'w')
os.chdir(path) #Change DIR
for i in os.listdir(path):
if i.endswith('.xml'):
z.write(i)
z.close()
Without changing DIR:
z = zipfile.ZipFile(zf, 'w')
for i in os.listdir(path):
if i.endswith('.xml'):
z.write(os.path.join(path, i))
z.close()

read contents of a file from a list of file with os.listdir() (python)

I need to read the contents of a file from the list of files from a directory with os.listdir. My working scriptlet is as follows:
import os
path = "/Users/Desktop/test/"
for filename in os.listdir(path):
with open(filename, 'rU') as f:
t = f.read()
t = t.split()
print(t)
print(t) gives me all the contents from all the files at once present in the directory (path).
But I like to print the contents on first file, then contents of the second and so on, until all the files are read from in dir.
Please guide ! Thanks.

You can print the file name.
Print the content after the file name.
import os
path = "/home/vpraveen/uni_tmp/temp"
for filename in os.listdir(path):
with open(filename, 'rU') as f:
t = f.read()
print filename + " Content : "
print(t)

First, you should find the path of each file using os.path.join(path, filename). Otherwise you'll loop wrong files if you change the variable path. Second, your script already provides the contents of all files starting with the first one. I added a few lines to the script to print the file path and an empty line to see where the contents end and begin:
import os
path = "/Users/Desktop/test/"
for filename in os.listdir(path):
filepath = os.path.join(path, filename)
with open(filepath, 'rU') as f:
content = f.read()
print(filepath)
print(content)
print()

os.listdir returns the name of the files only. you need to os.path.join that name with the path the files live in - otherwise python will look for them in your current working directory (os.getcwd()) and if that happens not to be the same as path python will not find the files:
import os
path = "/Users/Desktop/test/"
for filename in os.listdir(path):
print(filename)
file_path = os.path.join(path, filename)
print(file_path)
..
if you have pathlib at your disposal you can also:
from pathlib import Path
path = "/Users/Desktop/test/"
p = Path(path)
for file in p.iterdir():
if not file.is_file():
continue
print(file)
print(file.read_text())

Python file-IO and zipfile. Trying to loop through all the files in a folder and then loop through the texts in respective file using Python

Trying to extract all the zip files and giving the same name to the folder where all the files are gonna be.
Looping through all the files in the folder and then looping through the lines within those files to write on a different text file.
This is my code so far:
#!usr/bin/env python3
import glob
import os
import zipfile
zip_files = glob.glob('*.zip')
for zip_filename in zip_files:
dir_name = os.path.splitext(zip_filename)[0]
os.mkdir(dir_name)
zip_handler = zipfile.ZipFile(zip_filename, "r")
zip_handler.extractall(dir_name)
path = dir_name
fOut = open("Output.txt", "w")
for filename in os.listdir(path):
for line in filename.read().splitlines():
print(line)
fOut.write(line + "\n")
fOut.close()
This is the error that I encounter:
for line in filename.read().splitlines():
AttributeError: 'str' object has no attribute 'read'

You need to open the file and also join the path to the file, also using splitlines and then adding a newline to each line is a bit redundant:
path = dir_name
with open("Output.txt", "w") as fOut:
for filename in os.listdir(path):
# join filename to path to avoid file not being found
with open(os.path.join(path, filename)):
for line in filename:
fOut.write(line)
You should always use with to open your files as it will close them automatically. If the files are not large you can simply fOut.write(f.read()) and remove the loop.
You also set path = dir_name which means path will be set to whatever the last value of dir_name was in your first loop which may or may not be what you want. You can also use iglob to avoid creating a full list zip_files = glob.iglob('*.zip').

Extract files from zip without keeping the structure using python ZipFile?

I try to extract all files from .zip containing subfolders in one folder. I want all the files from subfolders extract in only one folder without keeping the original structure. At the moment, I extract all, move the files to a folder, then remove previous subfolders. The files with same names are overwrited.
Is it possible to do it before writing files?
Here is a structure for example:
my_zip/file1.txt
my_zip/dir1/file2.txt
my_zip/dir1/dir2/file3.txt
my_zip/dir3/file4.txt
At the end I whish this:
my_dir/file1.txt
my_dir/file2.txt
my_dir/file3.txt
my_dir/file4.txt
What can I add to this code ?
import zipfile
my_dir = "D:\\Download\\"
my_zip = "D:\\Download\\my_file.zip"
zip_file = zipfile.ZipFile(my_zip, 'r')
for files in zip_file.namelist():
zip_file.extract(files, my_dir)
zip_file.close()
if I rename files path from zip_file.namelist(), I have this error:
KeyError: "There is no item named 'file2.txt' in the archive"

This opens file handles of members of the zip archive, extracts the filename and copies it to a target file (that's how ZipFile.extract works, without taking care of subdirectories).
import os
import shutil
import zipfile
my_dir = r"D:\Download"
my_zip = r"D:\Download\my_file.zip"
with zipfile.ZipFile(my_zip) as zip_file:
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 = open(os.path.join(my_dir, filename), "wb")
with source, target:
shutil.copyfileobj(source, target)

It is possible to iterate over the ZipFile.infolist(). On the returned ZipInfo objects you can then manipulate the filename to remove the directory part and finally extract it to a specified directory.
import zipfile
import os
my_dir = "D:\\Download\\"
my_zip = "D:\\Download\\my_file.zip"
with zipfile.ZipFile(my_zip) as zip:
for zip_info in zip.infolist():
if zip_info.filename[-1] == '/':
continue
zip_info.filename = os.path.basename(zip_info.filename)
zip.extract(zip_info, my_dir)

Just extract to bytes in memory,compute the filename, and write it there yourself,
instead of letting the library do it - -mostly, just use the "read()" instead of "extract()" method:
Python 3.6+ update(2020) - the same code from the original answer, but using pathlib.Path, which ease file-path manipulation and other operations (like "write_bytes")
from pathlib import Path
import zipfile
import os
my_dir = Path("D:\\Download\\")
my_zip = my_dir / "my_file.zip"
zip_file = zipfile.ZipFile(my_zip, 'r')
for files in zip_file.namelist():
data = zip_file.read(files, my_dir)
myfile_path = my_dir / Path(files.filename).name
myfile_path.write_bytes(data)
zip_file.close()
Original code in answer without pathlib:
import zipfile
import os
my_dir = "D:\\Download\\"
my_zip = "D:\\Download\\my_file.zip"
zip_file = zipfile.ZipFile(my_zip, 'r')
for files in zip_file.namelist():
data = zip_file.read(files, my_dir)
# I am almost shure zip represents directory separator
# char as "/" regardless of OS, but I don't have DOS or Windos here to test it
myfile_path = os.path.join(my_dir, files.split("/")[-1])
myfile = open(myfile_path, "wb")
myfile.write(data)
myfile.close()
zip_file.close()

A similar concept to the solution of Gerhard Götz, but adapted for extracting single files instead of the entire zip:
with ZipFile(zipPath, 'r') as zipObj:
zipInfo = zipObj.getinfo(path_in_zip))
zipInfo.filename = os.path.basename(destination)
zipObj.extract(zipInfo, os.path.dirname(os.path.realpath(destination)))

In case you are getting badZipFile error. you can unzip the archive using 7zip sub process. assuming you have installed the 7zip then use the following code.
import subprocess
my_dir = destFolder #destination folder
my_zip = destFolder + "/" + filename.zip #file you want to extract
ziploc = "C:/Program Files/7-Zip/7z.exe" #location where 7zip is installed
cmd = [ziploc, 'e',my_zip ,'-o'+ my_dir ,'*.txt' ,'-r' ]
#extracting only txt files and from all subdirectories
sp = subprocess.Popen(cmd, stderr=subprocess.STDOUT, stdout=subprocess.PIPE)