CODEWITHSUNDEEP
pythonwindowscompression
lodkkx
lodkkx

Reputation: 1173

Extracting zip file contents to specific directory in Python 2.7

This is the code I am currently using to extract a zip file that lives in the same current working directory as the script. How can I specify a different directory to extract to?

The code I tried is not extracting it where I want.

import zipfile

fh = open('test.zip', 'rb')
z = zipfile.ZipFile(fh)
for name in z.namelist():
    outfile = open(name, 'wb')
    outfile.write('C:\\'+z.read(name))
    outfile.close()
fh.close()

Upvotes: 54

Views: 115251

Answers (6)

secretmike
secretmike

Reputation: 9884

I think you've just got a mixup here. Should probably be something like the following:

import zipfile

fh = open('test.zip', 'rb')
z = zipfile.ZipFile(fh)
for name in z.namelist():
    outpath = "C:\\"
    z.extract(name, outpath)
fh.close()

and if you just want to extract all the files:

import zipfile

with zipfile.ZipFile('test.zip', "r") as z:
    z.extractall("C:\\")

Use pip install zipfile36 for recent versions of Python

import zipfile36

Upvotes: 114

Peter Gibson
Peter Gibson

Reputation: 19564

If you just want to extract a zip file from the command line using Python (say because you don't have the unzip command available), then you can call the zipfile module directly

python -m zipfile -e monty.zip target-dir/

Take a look at the docs. It also supports compression and listing the contents.

Upvotes: 4

MD SARFARAZ
MD SARFARAZ

Reputation: 117

I've modified the code to ask the user input the file name and its path where it need to be extracted and so the user will've more control on where to put the extracted folder and what name should be assigned to the extracted folder.

import zipfile

#picking zip file from the directory
ZipFileName = raw_input("Enter full path to zip file:")  
fh = open( ZipFileName , 'rb')
z = zipfile.ZipFile(fh)

#assigning a name to the extracted zip folder
DestZipFolderName = raw_input("Assign destination folder a name: ")
DestPathName = raw_input("Enter destination directory: ")
DestPath = DestPathName + "\\" + DestZipFolderName

for name in z.namelist():   
    outpath = DestPath
    z.extract(name, outpath)
fh.close()

Upvotes: 0

Slakker
Slakker

Reputation: 171

Adding to secretmike's answer above with support for python 2.6 for extracting all files.

import zipfile
import contextlib


with contextlib.closing(zipfile.ZipFile('test.zip', "r")) as z:
   z.extractall("C:\\")

Upvotes: 5

fiatjaf
fiatjaf

Reputation: 12167

I tried the other answers in this thread, but the final solution for me was simply:

zfile = zipfile.ZipFile('filename.zip')
zfile.extractall(optional_target_folder)

Look at extractall, but use it only with trustworthy zip files.

Upvotes: 12

ratmatz
ratmatz

Reputation: 569

Peter de Rivaz has a point in the comment above. You are going to want to have the directory in the call to open(). You are going to want to do something like this:

import zipfile
import os

os.mkdir('outdir')
fh = open('test.zip','rb')
z = zipfile.ZipFile(fh)
for name in z.namelist():
    outfile = open('outdir'+'/'+name, 'wb')
    outfile.write()
    outfile.close()
fh.close()

Upvotes: 2

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