python 7z extraction with subprocess

Question

I try extract my archive using the subprocess:

subprocess.call(['7z', 'x', '-r', '-y', '-o %s' % os.path.normpath("C:/temp"), archivePath], shell = True)

but I get an error:

7-Zip [64] 9.20 Copyright (c) 1999-2010 Igor Pavlov 2010-11-18

Processing archive: \172.16.0.30\TestFarm\testdata\testdata.7z

Error: Can not create output directory C:\temp\

System error: The filename, directory name, or volume label syntax is incorrect.

2

How can I do it? Why it happens? If I use command line console it work perfect.

Answer 1

Set shell=False .

Set the output directory to be '-o%s' % directory.

You are prepending a space before the directory on the 7z command line.