Python / subprocess> Symbolic link Implementation to get only the destination file name> os.readlink ()

Operating environment


Xeon E5-2620 v4 (8 cores) x 2
32GB RAM
CentOS 6.8 (64bit)
openmpi-1.8.x86_64 and its-devel
mpich.x86_64 3.1-5.el6 and its-devel
gcc version 4.4.7 (And gfortran)
NCAR Command Language Version 6.3.0
WRF v3.7.Use 1.
Python 3.6.0 on virtualenv

$ touch test_original_170324
$ ln -fs test_original_170324 LN_link_170324
$ ls -l LN_link_170324 
lrwxrwxrwx 1 user user 20 Mar 24 16:06 LN_link_170324 -> test_original_170324

I want to get the original link file of LN_link_170324, test_original_170324, from a Python script.

Reference Get the return value of an external shell script (ls) with python3 Reference https://docs.python.jp/3/library/subprocess.html

test_python_170324j.py


from subprocess import getoutput

# Python 3.6.0 on virtualenv

cmd = 'ls -l LN_link_170324'
last = getoutput(cmd).split(' ')[-1]
print(last)

result


$ python test_python_170324j.py 
test_original_170324

What you taught me

(Addition 2017/03/25)

@ komeda-shinji told me os.readlink () in Comment.

Thank you for the information.

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