How to get symlink target in Python?

Question

Using Python, I need to check whether hundreds of symlinks are correct and recreate them when not. What I do now is to compare real paths of what I want and what I have, but it's slow because it's over NFS with an automount.

Otherwise I'm going to run a subprocess with the command 'ls -l' and work on the list of strings returned. I would prefer a better solution, using a Python library...

Edit1: I have: link_name -> link_target and then link_target -> a_real_file. What I need is to extract link_target from link_name, not a_real_file. I don't care if the real file does not exist.

Edit2: Maybe I did not express correctly. What I mean by a correct symlink is 'a link that point to a predefined path, even if it does not exist'. So I need to check that:

link_name_1 -> target_1
link_name_2 -> target_2

That's why I need to extract targets, not the real paths. Then I compare them to a reference (dictionary). So my question is: How do I extract the target path?

wim · Accepted Answer · 2017-02-23 21:53:48Z

80

The problem with os.readlink() is it will only resolve 1 step of the link. We can have a situation where A links to another link B, and B link is dangling.

$ ln -s /tmp/example/notexist /tmp/example/B
$ ln -s /tmp/example/B /tmp/example/A
$ ls -l /tmp/example
A -> /tmp/example/B
B -> /tmp/example/notexist

Now in Python, os.readlink gives you the first target.

>>> import os
>>> os.readlink('A')
'/tmp/example/B'

But in most situations I assume we are interested in the resolved path. So pathlib can help here:

>>> from pathlib import Path
>>> Path('A').resolve()
PosixPath('/tmp/example/notexist')

For older Python versions:

>>> os.path.realpath('A')
'/tmp/example/notexist'

answered Feb 23, 2017 at 21:53

wim

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2 Comments

mxl Feb 6 at 8:50

let me tell you from my own experience that in Python 2.7 on Windows (at least in my environment -- PY 2.7.15 32bit, W 11 Pro 24H2 26100.2894), python 2.7 returns c:\.....\file.txt while python 3.10 returns E:\.....\file.txt for the same file which is inside a directory junction pointing to E: located on c:.

Hans Krupakar Jun 16 at 6:50

You can use a combination of islink and readlink for the resolved path in cases of dangling links if you want to stick to os. path = 'A'; while os.path.islink(path): path = os.path.readlink(path)

Aran-Fey · Accepted Answer · 2019-08-27 14:14:02Z

23

You need to look at os.readlink().

edited Aug 27, 2019 at 14:14

Aran-Fey

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answered Oct 20, 2015 at 11:41

Armin Rigo

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3 Comments

SuperBiasedMan Over a year ago

Please consider editing your post to add more explanation about what your code does and why it will solve the problem. An answer that mostly just contains code (even if it's working) usually wont help the OP to understand their problem.

zorggy Over a year ago

Thank you Armin, it is exactly what I needed. I've been able to get the target and the task is now far faster.

gerardw Over a year ago

The original poster's problem was they did not know os.readlink() existed. This answer totally solves that problem.

SuperNova · Accepted Answer · 2020-09-29 09:10:18Z

19

In Python 3.9 or above, pathlib.Path.readlink() can be used.

>>> p = Path('mylink')
>>> p.symlink_to('setup.py')
>>> p.readlink()
PosixPath('setup.py')

edited Sep 29, 2020 at 9:10

answered Sep 29, 2020 at 8:25

SuperNova

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1 Comment

Tomerikoo Over a year ago

I think everyone should stop using os.path and move to pathlib where possible. Nice work updating old questions with new features. +1 for hopefully this will reach upwards :)

AChampion · Accepted Answer · 2022-05-21 16:25:25Z

3

To determine if a link is broken, you can, os.walk and test os.path.exists(path) which will return False for a broken link. You can then use os.path.realpath(path) to find out what the link is supposed to be pointing to.
Something like (untested):

for root, dirs, files in os.walk('<path>'):
    for file in files:
         f = os.path.join(root, file)
         if os.path.islink(f) and not os.path.exists(f):
             print("Broken: {} -> {}".format(f, os.path.realpath(f)))

edited May 21, 2022 at 16:25

answered Oct 20, 2015 at 9:18

AChampion

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2 Comments

antred Over a year ago

Beware that os.path.realpath() does not currently work on Windows (bugs.python.org/issue9949).

Nathanael Skrepek Over a year ago

For some reason I cannot edit this post. So I just write my edits as a comment: the method os.join in the 3rd line does not exists it has to be os.path.join.

user2879704 · Accepted Answer · 2015-10-20 09:08:18Z

-3

To determine if a directory entry is a symlink use this:

os.path.islink(path)

Return True if path refers to a directory entry that is a symbolic link. Always False if symbolic links are not supported.

answered Oct 20, 2015 at 9:08

user2879704

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