31

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.

Niels B.
  • 445
  • 1
  • 4
  • 8

5 Answers5

42

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.

Achilleas
  • 536
  • 4
  • 6
  • *"...helps to **imagine** that the created symlink simply holds text...."* Isn't this the literal truth? – Wildcard Feb 16 '16 at 00:00
  • 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 01 '17 at 08:23
  • 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 01 '17 at 10:35
  • 3
    In linux (gnu ln) the man page calls the first argument `target` and the second `link` (http://man7.org/linux/man-pages/man1/ln.1.html). But in BSD (including OS X) the first is called `source` and the second `target` (https://www.freebsd.org/cgi/man.cgi?ln). Quite confusing. – cristoper Jun 15 '18 at 15:56
8

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

ln -sr foo ~/bin/foo_link
Geremia
  • 1,163
  • 1
  • 13
  • 23
  • 6
    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 00:11
4

To save some typing, you can do 

ln -s "$PWD"/foo ~/bin/foo_link
Jeff Schaller
  • 66,199
  • 35
  • 114
  • 250
cosine
  • 76
  • 2
2

Use realpath

cd /Users/niels/something
ln -s "$(realpath foo)" ~/bin/foo_link
Kusalananda
  • 320,670
  • 36
  • 633
  • 936
MortezaE
  • 121
  • 4
-1

How about:

 $ cd /Users/niels/something
 $ ln -s ./foo ~/bin/foo_link
ptierno
  • 202
  • 1
  • 4