CODEWITHSUNDEEP
python
Ian
Ian

Reputation: 1717

Call 7zip from Python on Windows

I'm trying to figure out how to invoke 7zip, on Windows, from a Python program.

I'm trying:

 stdout = subprocess.run(['C:\\Program Files\\7-Zip\\7z.exe', "a -t7z -mx0", "C:\\Users\\IanWo\\test.7z", "C:\\Users\\IanWo\\test.txt", "C:\\Users\\IanWo\\test2.txt"], shell=True, check=True, capture_output=True, text=True).stdout
 print(stdout)

but am getting:

Traceback (most recent call last):
File "D:\Normal Backup\Code\ProcessRetrospectBackups\process.py", line 93, in <module>
    stdout = subprocess.run(['C:\\Program Files\\7-Zip\\7z.exe', "a -t7z -mx0", "C:\\Users\\IanWo\\test.7z", "C:\\Users\\IanWo\\test.txt", "C:\\Users\\IanWo\\test2.txt"], shell=True, check=True, capture_output=True, text=True).stdout

  File "C:\Users\IanWo\AppData\Local\Programs\Python\Python39\lib\subprocess.py", line 524, in run
    raise CalledProcessError(retcode, process.args,
subprocess.CalledProcessError: Command '['C:\\Program Files\\7-Zip\\7z.exe', 'a -t7z -mx0', 'C:\\Users\\IanWo\\test.7z', 'C:\\Users\\IanWo\\test.txt', 'C:\\Users\\IanWo\\test2.txt']' returned non-zero exit status 7.
>>>

It invokes 7z.exe fine as long as I don't have any arguments. I've tried with and without shell=True with no change.

Upvotes: 0

Views: 641

Answers (1)

Tim Roberts
Tim Roberts

Reputation: 54917

@ThiefMaster is right. Here's the correct call:

stdout = subprocess.run(['C:\\Program Files\\7-Zip\\7z.exe', "a", "-t7z", "-mx0", "C:\\Users\\IanWo\\test.7z", "C:\\Users\\IanWo\\test.txt", "C:\\Users\\IanWo\\test2.txt"], shell=True, check=True, capture_output=True, text=True).stdout
print(stdout)

Upvotes: 2

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