CODEWITHSUNDEEP
pythonpython-3.xsubprocesspopenls
M P Mathugama
M P Mathugama

Reputation: 1418

List only files and directory names from subprocess ls -l

In python subprocess using Popen or check_output, I need to list files and directories in a given source directory. But I can only use the command ls -l.

Sample code

cmd = ["ls", "-l", source]
proc = subprocess.Popen(cmd, stdout=subprocess.PIPE)
stdout, stderr = proc.communicate()
exitcode = proc.returncode

if exitcode != 0:
    raise SystemError("Exitcode '{}', stderr: '{}', stdout: '{}'  for command: '{}'".format(
        exitcode, stderr, stdout, cmd))

From above proc, by using grep or any other way, can I get only a list of files and directory names inside source directory without other information?

Upvotes: 1

Views: 5510

Answers (3)

user3064538
user3064538

Reputation:

Parsing the output of ls is a bad idea for a few reasons. If your file name has a trailing space, then ls will display it as 'trailing space ' and if you try to open("'trailing space '") it won't work. Also file names can contain newlines.

Use pathlib instead:

from pathlib import Path
source = Path("/path/to/some/directory")
[x.name for x in source.iterdir()]
# ['a_file', 'some_other_file.txt', 'a_directory']

Upvotes: 2

bart cubrich
bart cubrich

Reputation: 1254

As Charles Duffy mentioned, you can use os. Like this.

import os
directory=#wherever you want to search
files_and_directories=os.listdir(directory)
Out: ['Directories and file names in a list']

Upvotes: 1

Nihal Sangeeth
Nihal Sangeeth

Reputation: 5535

If you insist on using subprocess please try:

[x.split(' ')[-1] for x in stdout.decode().split('\n')[1:-1]]

Obviously this is a pretty "hacky" way of doing this. Instead I can suggest the standard library glob

import glob
glob.glob(source + '/*')

returns a list of all file/directory names in source.

Edit:

cmd = ["ls", source]
proc = subprocess.Popen(cmd, stdout=subprocess.PIPE)
stdout, stderr = proc.communicate()
exitcode = proc.returncode
stdout.decode("utf-8").split('\n')[:-1]

Should also do it. -l option is not necessary here.

Upvotes: 3

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