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Python doesn't recognize zip files as zip files

I iterate through the directories and want to find all zip files and add them to download_all.zip
I am sure there are zip files, but Python doesn't recognize those zip files as zip files. Why is that?
my code:
os.chdir(boardpath)
# zf = zipfile.ZipFile('download_all.zip', mode='w')
z = zipfile.ZipFile('download_all.zip', 'w') #creating zip download_all.zip file
for path, dirs, files in os.walk(boardpath):
for file in files:
print file
if file.endswith('.zip'): # find all zip files
print ('adding', file)
z.write(file) # error shows: doesn't file is a str object, not a zip file
z.close()
z = zipfile.ZipFile("download_all.zip")
z.printdir()
I tried:
file.printdir()
# I got the following error: AttributeError: 'str' object has no attribute 'printdir'

zipfile.Zipfile.write(name), name actually stands for full file path, not just filename.
import os #at the top
if file.endswith('.zip'): # find all zip files
filepath = os.path.join(path, file)
print ('adding', filepath)
z.write(filepath) # no error

As stated in the ZipFile.write's doc, the filename argument must be relative to the archive root. So the following line:
z.write(file)
Should be:
z.write(os.path.relpath(os.path.join(path, file)))

The files that os/walk() yields are lists of filenames. These filenames are just strings (which don't have a printdir() method).

You want to use the context management while opening up the zip file archive and writing to it for each file that you find, hence the use of with. In addition, since you're walking through a directory structure, you need to full qualify each file's path.
import os
import Zipfile
with zipfile.ZipFile('download_all.zip', 'w') as zf:
for path, dirs, files in os.walk('/some_path'):
for file in files:
if file.endswith('.zip'):
zf.write(os.path.join(path, file))

How to read particular text files out-of multiple files in a sub directories in python

I have a one folder, within it contains 5 sub-folders.
Each sub folder contains some 'x.txt','y.txt' and 'z.txt' files and it repeats in every sub-folders
Now I need to read and print only 'y.txt' file from all sub-folders.
My problem is I'm unable to read and print 'y.txt' files. Can you tell me how solve this problem.
Below is my code which I have written for reading y.txt file
import os, sys
import pandas as pd
file_path = ('/Users/Naga/Desktop/Python/Data')
for root, dirs, files in os.walk(file_path):
for name in files:
print(os.path.join(root, name))
pd.read_csv('TextInformation.txt',delimiter=";", names = ['Name', 'Value'])
error :File TextInformation.txt does not exist: 'TextInformation.txt'

You could also try the following approach to fetch all y.txt files from your subdirectories:
import glob
import pandas as pd
# get all y.txt files from all subdirectories
all_files = glob.glob('/Users/Naga/Desktop/Python/Data/*/y.txt')
for file in all_files:
data_from_this_file = pd.read_csv(file, sep=" ", names = ['Name', 'Value'])
# do something with the data
Subsequently, you can apply your code to all the files within the list all_files. The great thing with glob is that you can use wilcards (*). Using them you don't need the names of the subdirectories (you can even use it within the filename, e.g. *y.txt). Also see the documentation on glob.

Your issue is forgot adding the parent path of 'y.txt' file. I suggest this code for you, hope it help.
import os
pth = '/Users/Naga/Desktop/Python/Data'
list_sub = os.listdir(pth)
filename = 'TextInformation.txt'
for sub in list_sub:
TextInfo = open('{}/{}/{}'.format(pth, sub, filename), 'r').read()
print(TextInfo)

I got you a little code. you can personalize it anyway you like but the code works for you.
import os
for dirPath,foldersInDir,fileName in os.walk(path_to_main_folder):
if fileName is not []:
for file in fileName:
if file.endswith('y.txt'):
loc = os.sep.join([dirPath,file])
y_txt = open(loc)
y = y_txt.read()
print(y)
But keep in mind that {path_to_main} is the path that has the subfolders.

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()

Extracting a single file from a zip archive without iterating over the entire name list in Python

I have a zip file with a folder in it, like this:
some.zip/
some_folder/
some.xml
...
I am using the zipfile library.
What I want is to open only some.xml file, but I don't now the some_folder name.
My solution looks like this:
def get_xml(zip_file):
for filename in zip_file.namelist():
if filename.endswith('some.xml'):
return zip_file.open(filename)
I would like to know if there is a better solution other than scanning the entire list.

This prints the list of directories inside the test.zip file:
from zipfile import ZipFile
with ZipFile('test.zip', 'r') as f:
directories = [item for item in f.namelist() if item.endswith('/')]
print directories
If you know that there is only one directory inside, just take the first item: directories[0].

Do you want to get directory that containing some.xml?
import os
import zipfile
with zipfile.ZipFile('a.zip', 'r') as zf:
for name in zf.namelist():
if os.path.basename(name) == 'some.xml':
print os.path.dirname(name)

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)