is there equal syntax to "dir /s /b"

is there equal syntax to "dir /s /b" - Printable Version

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is there equal syntax to "dir /s /b" - kucingkembar - Aug-16-2022

sorry for my bad English,
currently i looking all files in a folder and subfolder like "dir /s /b" in windows-cmd

D:\test>dir /s /b
D:\test\test
D:\test\test (2).py
D:\test\test.py
D:\test\test\test.py

is there any equal syntax like this in python?
Note:
I know code

import os
for text in os.listdir():
    print(text)

but that only show files name, I need full path

RE: is there equal syntax to "dir /s /b" - Pedroski55 - Aug-16-2022

]Try glob??

import glob

path = '/home/pedro/summer2021/OMR/'

files = glob.glob(path + '/**', recursive=True)

for f in files:
    # shows all paths and files in path
    print(f)

or like this:

from pathlib import Path
mydir = Path(path)
path2python = '/home/pedro/summer2021/OMR/python/'
mydir = Path(path2python)
filelist = [filename for filename in mydir.iterdir() if filename.is_file()]

for filename in filelist:
    print(f"\nfilename: {filename.name}")
    print(f"file suffix: {filename.suffix}")
    print(f"full path: {filename.resolve()}")
    print(f"filepath parts: {filename.parts}")

import os
for root,dirs,files in os.walk(path):
    print('root is:', root)
    print('dirs is:', dirs)
    print('files are:', files)
    for file in files:
        print(os.path.join(root, file))

If I remember correctly, I got all this from snippsat, so please say thank you to him, not me!

RE: is there equal syntax to "dir /s /b" - kucingkembar - Aug-16-2022

(Aug-16-2022, 07:54 AM)Pedroski55 Wrote: ]Try glob??

import glob

path = '/home/pedro/summer2021/OMR/'

files = glob.glob(path + '/**', recursive=True)

for f in files:
    # shows all paths and files in path
    print(f)

or like this:

from pathlib import Path
mydir = Path(path)
path2python = '/home/pedro/summer2021/OMR/python/'
mydir = Path(path2python)
filelist = [filename for filename in mydir.iterdir() if filename.is_file()]

for filename in filelist:
    print(f"\nfilename: {filename.name}")
    print(f"file suffix: {filename.suffix}")
    print(f"full path: {filename.resolve()}")
    print(f"filepath parts: {filename.parts}")

import os
for root,dirs,files in os.walk(path):
    print('root is:', root)
    print('dirs is:', dirs)
    print('files are:', files)
    for file in files:
        print(os.path.join(root, file))

If I remember correctly, I got all this from snippsat, so please say thank you to him, not me!

thank you Pedroski55. i will give you reputation point