How to read contents of 7z file using python

Question

How can I read and save the contents of 7z. I use Python 2.7.9, I can extract or Archive like this, but I can't read contents in python, I only listing the file's contents in CMD

import subprocess
import os

source = 'filename.7z'
directory = 'C:\Directory'
pw = '123456'
subprocess.call(r'"C:\Program Files (x86)\7-Zip\7z.exe" x '+source +' -o'+directory+' -p'+pw)

Answer 1

You can use either libarchive or pylzma. If you can upgrade to python3.3+ you can use lzma, which is in the standard library.

Answer 2

I ended up in this situation where I was forced to use 7z, and also needed to know exactly which files were extracted from each zip archive. To deal with this, you can check the output of the call to 7z and look for the filenames. Here's what the output of 7z looks like:

$ 7z l sample.zip

7-Zip [64] 16.02 : Copyright (c) 1999-2016 Igor Pavlov : 2016-05-21
p7zip Version 16.02 (locale=utf8,Utf16=on,HugeFiles=on,64 bits,8 CPUs x64)

Scanning the drive for archives:
1 file, 472 bytes (1 KiB)

Listing archive: sample.zip

--
Path = sample.zip
Type = zip
Physical Size = 472

   Date      Time    Attr         Size   Compressed  Name
------------------- ----- ------------ ------------  ------------------------
2018-12-01 17:09:59 .....            0            0  sample1.txt
2018-12-01 17:10:01 .....            0            0  sample2.txt
2018-12-01 17:10:03 .....            0            0  sample3.txt
------------------- ----- ------------ ------------  ------------------------
2018-12-01 17:10:03                  0            0  3 files

and how to parse that output with python:

import subprocess

def find_header(split_line):
    return 'Name' in split_line and 'Date' in split_line

def all_hyphens(line):
    return set(line) == set('-')

def parse_lines(lines):
    found_header = False
    found_first_hyphens = False
    files = []
    for line in lines:

        # After the header is a row of hyphens
        # and the data ends with a row of hyphens
        if found_header:
            is_hyphen = all_hyphens(''.join(line.split()))

            if not found_first_hyphens:
                found_first_hyphens = True
                # now the data starts
                continue

            # Finding a second row of hyphens means we're done
            if found_first_hyphens and is_hyphen:
                return files

        split_line = line.split()

        # Check for the column headers
        if find_header(split_line):
            found_header=True
            continue

        if found_header and found_first_hyphens:
            files.append(split_line[-1])
            continue

    raise ValueError("We parsed this zipfile without finding a second row of hyphens")



byte_result=subprocess.check_output('7z l sample.zip', shell=True)
str_result = byte_result.decode('utf-8')
line_result = str_result.splitlines()
files = parse_lines(line_result)

Answer 3

Shelling out and calling 7z will extract files and then you can access those files using standard File Access calls (I don't know Python - but it must be able to access files!).

If you want to look inside a 7z archive directly within Python, then you'll need to use a library. Here's one: https://pypi.python.org/pypi/libarchive - I can't vouch for it as I said - I'm not a Python user - but using a 3rd party library is usually pretty easy in all languages.

Generally, 7z Support seems limited. If you can use alternative formats (zip/gzip) then I think you'll find the range of Python libraries (and example code) is more comprehensive.

Hope that helps.