26

I'm trying to create a bunch of symbolic links, but I can't figure out why this is working

ln -s /Users/niels/something/foo ~/bin/foo_link

while this

cd /Users/niels/something
ln -s foo ~/bin/foo_link

is not.

I believe it has something to do with foo_link linking to foo in /Users/niels/bin instead of /Users/niels/something

So the question is, how do I create a symbolic link that points to an absolute path, without actually typing it?

For reference, I am using Mac OS X 10.9 and Zsh.

33

The easiest way to link to the current directory as an absolute path, without typing the whole path string would be

ln -s "$(pwd)/foo" ~/bin/foo_link

The target (first) argument for the ln -s command works relative to the symbolic link's location, not your current directory. It helps to know that, essentially, the created symlink (the second argument) simply holds the text you provide for the first argument.

Therefore, if you do the following:

cd some_directory
ln -s foo foo_link

and then move that link around

mv foo_link ../some_other_directory
ls -l ../some_other_directory

you will see that foo_link tries to point to foo in the directory it is residing in. This also works with symbolic links pointing to relative paths. If you do the following:

ln -s ../foo yet_another_link

and then move yet_another_link to another directory and check where it points to, you'll see that it always points to ../foo. This is the intended behaviour, since many times symbolic links might be part of a directory structure that can reside in various absolute paths.

In your case, when you create the link by typing

ln -s foo ~/bin/foo_link

foo_link just holds a link to foo, relative to its location. Putting $(pwd) in front of the target argument's name simply adds the current working directory's absolute path, so that the link is created with an absolute target.

  • "...helps to imagine that the created symlink simply holds text...." Isn't this the literal truth? – Wildcard Feb 16 '16 at 0:00
  • 1
    It is. Perhaps I could change imagine to "know" or "understand". – Achilleas Feb 17 '16 at 13:21
  • Perhaps I have gotten myself confused (again). It seems where you say "target" you mean "source". The source_file (first) argument is where the link points, which is verbatim what you enter on the command. The target_file (second) argument becomes the name of the link unless what you enter is a directory, in which case the link name is the same as the basename of the source_file but placed in the directory target_file. – Charlie Gorichanaz Apr 1 '17 at 8:23
  • Y̶o̶u̶'̶r̶e̶ ̶r̶i̶g̶h̶t̶,̶ ̶t̶h̶e̶r̶e̶'̶s̶ ̶a̶ ̶m̶i̶s̶t̶a̶k̶e̶ ̶i̶n̶ ̶t̶h̶e̶ ̶a̶n̶s̶w̶e̶r̶.̶ ̶C̶r̶a̶z̶y̶ ̶h̶o̶w̶ ̶i̶t̶ ̶d̶i̶d̶n̶'̶t̶ ̶g̶e̶t̶ ̶c̶a̶u̶g̶h̶t̶ ̶s̶o̶o̶n̶e̶r̶.̶ ̶F̶i̶x̶i̶n̶g̶.̶ No, the answer was right. By the ln manpage, target is the first argument (where the link points to) whereas the second argument is simply referred to as the link (source is not mentioned). – Achilleas Apr 1 '17 at 10:35
  • 3
    In linux (gnu ln) the man page calls the first argument target and the second link (man7.org/linux/man-pages/man1/ln.1.html). But in BSD (including OS X) the first is called source and the second target (freebsd.org/cgi/man.cgi?ln). Quite confusing. – cristoper Jun 15 '18 at 15:56
5

Using the -r (--relative) flag will make this work:

ln -sr foo ~/bin/foo_link
  • 4
    Beware that -r is a GNUism, i.e. non POSIX so won't work in the OP case as the standard OS X ln command is BSD based. – jlliagre Feb 16 '16 at 0:11
2

To save some typing, you can do

ln -s "$PWD"/foo ~/bin/foo_link
  • @Freddy, it depends on the shell. In fish or zsh, it would be OK. In Bourne-like shells including bash and with the default value of $IFS, SPC would indeed be a problem, but also TAB, NL and all globbing characters (at least *, ? and [...]). – Stéphane Chazelas Sep 3 at 19:55
  • @StéphaneChazelas Yes, you're right and the question is tagged with bash, so it's a generally a good idea to quote it. My bad, I should have been more specific. – Freddy Sep 3 at 20:03
0

How about:

 $ cd /Users/niels/something
 $ ln -s ./foo ~/bin/foo_link
  • 1
    Nope, doesn't do the trick. – Niels B. Apr 16 '14 at 23:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.