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I have zip files in each of my folders

>>> subprocess.run(["ls", "-l"])
total 60
drwxrwxr-x 2 milenko milenko 4096 јул  4 13:03 Chapter01
drwxrwxr-x 2 milenko milenko 4096 јул  4 13:03 Chapter02
drwxrwxr-x 2 milenko milenko 4096 јул  4 13:03 Chapter03
drwxrwxr-x 2 milenko milenko 4096 јул  4 13:03 Chapter04
drwxrwxr-x 2 milenko milenko 4096 јул  4 13:03 Chapter05
drwxrwxr-x 2 milenko milenko 4096 јул  4 13:03 Chapter06
drwxrwxr-x 2 milenko milenko 4096 јул  4 13:03 Chapter07
drwxrwxr-x 2 milenko milenko 4096 јул  4 13:03 Chapter08
drwxrwxr-x 2 milenko milenko 4096 јул  4 13:03 Chapter09
drwxrwxr-x 2 milenko milenko 4096 јул  4 13:03 Chapter10
drwxrwxr-x 2 milenko milenko 4096 јул  4 13:03 Chapter11
drwxrwxr-x 2 milenko milenko 4096 јул  4 13:03 Chapter12
drwxrwxr-x 2 milenko milenko 4096 јул  4 13:03 Chapter13
-rw-rw-r-- 1 milenko milenko 1062 јул  4 13:03 LICENSE
-rw-rw-r-- 1 milenko milenko 2415 јул  4 13:03 README.md
CompletedProcess(args=['ls', '-l'], returncode=0)

What I really want is to unzip them with one subprocess command line.But what I have tried does not work

subprocess.run(['unzip','/home/milenko/Building-Django-2.0-Web-Applications/'])
unzip:  cannot find or open /home/milenko/Building-Django-2.0-Web-Applications/, /home/milenko/Building-Django-2.0-Web-Applications/.zip or /home/milenko/Building-Django-2.0-Web-Applications/.ZIP.
CompletedProcess(args=['unzip', '/home/milenko/Building-Django-2.0-Web-Applications/'], returncode=9)

How to change this?

MikiBelavista
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1 Answers1

1

In the unzip you are not mentioning the name of the file its taking empty by default *.zip should your job

subprocess.run(['unzip','/home/milenko/Building-Django-2.0-Web-Applications/*.zip'])

You can also refer to this:

Convert this to subprocess

find . -name "*.zip" | xargs -P 5 -I fileName sh -c 'unzip -o -d "$(dirname "fileName")/$(basename -s .zip "fileName")" "fileName"'

This will extract and store the file in the same directory

https://stackoverflow.com/a/2318189/9216722

arunp9294
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