如何使用'ls -l'命令和Python获取给定“路径”的内容

Question

I am using python and i want to get listing of all of the files /directories (not nested) at a given path. Meaning i need exact equivalent out put of "ls -l" command using python.

For eg at path /opt/test/ ls -l out put is shown below.

-rw-r--r--  1 user  qa-others  16715 Jan 16 13:38 file_2001161337
-rw-r--r--  1 user  qa-others  16715 Jan 16 13:46 file_2001161346
-rw-r--r--  1 user  qa-others  16715 Jan 16 13:54 file_2001161353

My python code is shown below.

print(subprocess.check_output(['ls', '-l']))

How can i pass the path value ie "/opt/temp" and get the same put of "ls -l" as shown above?

Answer 1

You are far, far better off using os.listdir , which essentially exactly what you want.

You can also use os.scandir if you need other information about these folders, though you'll need to filter to ensure you're only picking up folders:

[e for e in os.scandir() if e.isdir]

Each of these functions takes a path argument if you want to explicitly specify it, otherwise they run on the current directory.

Answer 2

You can use pathlib.Path() for this (Python >=3.4):

from pathlib import Path

source = Path('/opt/temp')

# Get all children
content = source.glob('*')

By default this will return an iterator of pathlib.Path objects (cast iterator to list if you need a visual check). Then you can programmatically access file attributes using pahtlib.Path.stat() .