[SOLVED] Retrieving the filename from a shell

[jehoshua]

Needing to find the latest filename in a specific directory. This is very basic but works

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# Importing required module import subprocess   # Using system() method to # execute shell commands   subprocess.Popen('ls ~/Documents/Backups  -Artls | tail -1', shell=True)

960 -rw-r--r-- 1 ******** ******** 981156 Nov 28 08:01 Australian-2024-11-28.kmy

How to I retrieve that filename, so that I can then run 'gunzip' on the file and create the uncompressed file ? The gunzip part should be easy, it is retrieving that filename ?? Obviously STDOUT in some form. Also, I have read that using "ls" is not the best method ?

[jehoshua]

Interesting - I tried the code at https://python-forum.io/thread-43593-pos...#pid182722 , modified it accordingly, and no output. Eventually it worked; the tilde (~) to signify 'home' path was not being recognised at all. The path name had to be explicitly defined.

1	my_command = 'ls /home/username/Documents/Backups  -Artls | tail -1'

works

1	my_command = 'ls ~/Documents/Backups  -Artls | tail -1'

doesn't work.

[Gribouillis]

Try this perhaps

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from collections import deque from pathlib import Path import subprocess as sp     def last_entry(directory):     d = Path(directory).expanduser()     proc = sp.Popen(["ls", str(d), "-Artls"], stdout=sp.PIPE, encoding="utf8")     dq = deque(proc.stdout, maxlen=1)     proc.wait()     return dq[-1] if dq else None     if __name__ == "__main__":     x = last_entry("~/tmp")     print(x)

Also you can simplify this by not using the 'rls' options in ls. Then you only take the first entry in the list instead of the last and get the file name directly

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from pathlib import Path import subprocess as sp     def last_entry(directory):     d = Path(directory).expanduser()     proc = sp.Popen(["ls", str(d), "-At", "-1"], stdout=sp.PIPE, encoding="utf8")     last = next(proc.stdout, None)     proc.stdout.close()     proc.wait()     return last     if __name__ == "__main__":     x = last_entry("~/tmp")     print(x.strip())

[jehoshua]

Great, thank you @Gribouillis . Both of those scripts work just fine.

The first returns

960 -rw-r--r-- 1 ******** ******** 981156 Nov 28 19:16 Australian-2024-11-28.kmy

and the second returns

Australian-2024-11-28.kmy

so I can now use the second one as the filename to "gunzip"