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Python - Working with ZIP archives
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  Opening Zip Files

To start, import the zipfile module, and set the filename.

import zipfile
filename = 'zipfile.zip'

Working with zip archives is very similar to working with files (opens new window), you create the object by opening the zipfile, which lets you work on it before closing the file up again.

zip = zipfile.ZipFile(filename)
print(zip)
# <zipfile.ZipFile object at 0x0000000002E51A90>
zip.close()

In Python 2.7 and in Python 3 versions higher than 3.2, we can use the with context manager. We open the file in "read" mode, and then print a list of filenames:

with zipfile.ZipFile(filename, 'r') as z:
    print(zip)
    # <zipfile.ZipFile object at 0x0000000002E51A90>
  Examining Zipfile Contents

There are a few ways to inspect the contents of a zipfile. You can use the printdir to just get a variety of information sent to stdout

with zipfile.ZipFile(filename) as zip:
    zip.printdir()

    # Out:
    # File Name                                             Modified             Size
    # pyexpat.pyd                                    2016-06-25 22:13:34       157336
    # python.exe                                     2016-06-25 22:13:34        39576
    # python3.dll                                    2016-06-25 22:13:34        51864
    # python35.dll                                   2016-06-25 22:13:34      3127960
    # etc.

We can also get a list of filenames with the namelist method. Here, we simply print the list:

with zipfile.ZipFile(filename) as zip:
    print(zip.namelist())

# Out: ['pyexpat.pyd', 'python.exe', 'python3.dll', 'python35.dll', ... etc. ...]

Instead of namelist, we can call the infolist method, which returns a list of ZipInfo objects, which contain additional information about each file, for instance a timestamp and file size:

with zipfile.ZipFile(filename) as zip:
    info = zip.infolist()
    print(zip[0].filename)
    print(zip[0].date_time)
    print(info[0].file_size)

# Out: pyexpat.pyd
# Out: (2016, 6, 25, 22, 13, 34)
# Out: 157336
  Extracting zip file contents to a directory

Extract all file contents of a zip file

import zipfile
with zipfile.ZipFile('zipfile.zip','r') as zfile:
    zfile.extractall('path')

If you want extract single files use extract method, it takes name list and path as input parameter

import zipfile
f=open('zipfile.zip','rb')
zfile=zipfile.ZipFile(f)
for cont in zfile.namelist():
    zfile.extract(cont,path)
  Creating new archives

To create new archive open zipfile with write mode.

import zipfile
new_arch=zipfile.ZipFile("filename.zip",mode="w")

To add files to this archive use write() method.

new_arch.write('filename.txt','filename_in_archive.txt') #first parameter is filename and second parameter is filename in archive by default filename will taken if not provided
new_arch.close()

If you want to write string of bytes into the archive you can use writestr() method.

str_bytes="string buffer"
new_arch.writestr('filename_string_in_archive.txt',str_bytes)
new_arch.close()

Syntax

  • import zipfile
  • class zipfile.ZipFile(file, mode='r', compression=ZIP_STORED, allowZip64=True)

Remarks

If you try to open a file that is not a ZIP file, the exception zipfile.BadZipFile is raised.

In Python 2.7, this was spelled zipfile.BadZipfile, and this old name is retained alongside the new one in Python 3.2+

 

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