问题
So I am trying to encrypt a directory using python and I'm not sure what the best way to do that is. I am easily able to turn the folder into a zip file, but from there I have tried looking up how to encrypt it with AES, but couldn't get that to work and I have also tried encrypting using 7zip to archive the folder, but also couldn't get that to work, so if anybody has another solution to encrypt a directory or could point me in the right direction on how to use one of the previous methods that would be helpful. (I'm on windows if that has any significance)
回答1:
I still recommend 7-zip.
let's say you want to name the zip folder as myzip.zip
Import subprocess
zp = subprocess.call(['7z', 'a', 'your password', '-y', 'myzip.zip'] + ['your file'])
An alternative way:
Import pyminzip
level=4 #level of compression
pyminizip.compress("your file", "myzip.zip", "your password", level)
回答2:
Using 7-Zip through the subprocess module works. Here are some issues I encountered and had to resolve: You need to specify the path to 7zip separate from the cmd variable in the Popen subprocess, and build the command with variables rather than a solid string:
appPath="C:\Program Files\\7-Zip"
zApp="7z.exe"
zAction='a'
zPass='-pPASSWORD'
zAnswer='-y'
zDir=directoryToZip
progDir=os.path.join(appPath,zApp)
cmd = [zApp, zAction, zipFileName, zPass, zAnswer, zDir]
subprocess.Popen(cmd, executable=progDir, stderr=subprocess.STDOUT, stdout=subprocess.PIPE)
That will create a zip file (in the location with the name in the zipFileName variable), including the contents (directories and files) inside the "directoryToZip" path
progDir has to be specified for separate from the application you are calling as part of the Open command (this is the executable path), and the command string needed to be built out as variables to deal with the windows backslash escaping setup.
来源：https://stackoverflow.com/questions/45094647/encrypt-folder-or-zip-file-using-python