There is a post on this topic already, but it does not have an explicit answer to the fundamental question which I am re-asking here:
How do you make 7zip commands from Python?
Attempting to use the subprocess module, I implemented the following which runs but does nothing (from what I can tell):
import subprocess
cmd = ['7z', 'a', '"Test.7z"', '"Test"', '-mx9']
subprocess.Popen(cmd, stderr=subprocess.STDOUT, stdout=subprocess.PIPE)
I know that the following 7zip command works, as I have tested on the Windows command line itself:
7z a "Test.7z" "Test" -mx9
How could I implement that simple 7zip command from Python?
To begin a session, open a terminal window. Invoke the version of 7Zip you are using by entering "7z" for P7Zip (7z.exe), or "7za" for 7Zip for Windows (7za.exe) to start either the P7-Zip or 7za application prior to entering commands.
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.
import subprocess
cmd = ['7z', 'a', 'Test.7z', 'Test', '-mx9']
sp = subprocess.Popen(cmd, stderr=subprocess.STDOUT, stdout=subprocess.PIPE)
You can wrap it as a function using the following:
import subprocess
def sevenzip(filename, zipname, password):
print("Password is: {}".format(password))
system = subprocess.Popen(["7z", "a", zipname, filename, "-p{}".format(password)])
return(system.communicate())
This definitely works as I've tried and tested it. If you want to tweak it i.e. to Extract files then you can use the following:
def extractfiles(zipname):
system = subprocess.Popen(["7z", "e", zipname])
return(system.communicate())
Give this a try and lemme know how you get on.
Bear in mind this is for Linux. In Windows, swap "7z" with "C:\Program Files\7-Zip\7z.exe" (i think that's the right location).
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