Is there any way I can create a link, that dynamically updates. My specific situation is, that I have multiple exercises (directories) and I want to link to the latest one:

│   exercise_01/
│   │    files ...
│   exercise_02/
│   │    files ...
│   exercise_03/
│   │    files ...
│   exercise_latest/ -> exercise_03/

so if I cd exercises/exercise_latest it always goes to the latest one. Adding a new directory exercise_04 would make the link point to that one instead.

The solutions I can come up with are:

  1. A cron job that runs once a minute and relinks if it detects a new directory.
  2. Not make it a link but a script that cds to the latest directory.
  3. Manually maintaining it

Neither solution is particularly elegant. 1. is really inefficient and potentially too slow. I cannot copy files into 2.. 3. defeats the purpose.

Versioned software releases do something similar. For example python3 always links to the latest Python 3 version. They however probably update every time a new version is installed.

Is there anything I can do that is more elegant than my ideas?

  • What about combining option 2 with creating another script that will copy files into the latest directory? That seems like the simplest and most robust solution. – davidmneedham Dec 10 '18 at 15:06
  • A fourth option is to incorporate the updating of exercise_latest into the creation of a new subdirectory, rather than trying to maintain it when you need to use it. – AFH Dec 10 '18 at 15:35

The way I solved this is creating a launchctl / launchd agent which can listen to files. This is macOS specific, but there are similar solutions for Linux.

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE plist PUBLIC -//Apple Computer//DTD PLIST 1.0//EN
<plist version="1.0">


This executes a Python script (.ln_monitor.py)

#!/usr/bin/env python3

import os
import re
from pathlib import Path
import time

DIR = Path(os.path.dirname(os.path.realpath(__file__)))
LINK_ROOTS = ['exercise']

def cprint(*args):
    if VERBOSE:

def main():
    for root in LINK_ROOTS:
        regex = r'^' + root + r'_\d+$'
        matches = [
            directory for directory in DIR.iterdir()
            if directory.is_dir() and
            re.match(regex, directory.name)]
        match_link = DIR / root
        if matches:
            match = sorted(matches)[-1]
            cprint(f'link to {match.name}')
            if match_link.is_symlink():
                if match_link.resolve() == match:
                    cprint('is already linked')
                cprint('no link yet')

if __name__ == '__main__':

the LINK_ROOTS here contain all the files that are to be linked. All files matching matching a link root with an underscore and any numbers behind will match, the greatest one will be linked to the link root.

so if LINK_ROOTS = ['exercise'] the greatest one will be linked like this exercise -> exercise_04.

This is not a perfect solution, but it works quite reliably and is not super inefficient.

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