How to find symlink target name in script
 

Hey,

I have a symlink
~/bin/foo/image.link
that points to
~/photos/image.jpg

I am looking for a command kind of like this:

But I want the name of the target file not just info about it. The 'file' command would work fine except that I don't know how to pull out just the filename for my script, grep with a regex just lists the whole line.

Is there a command you know of/option to 'ls' ?? Or is 'sed'/'awk' something I should look into??

Thanks,

file somefile | awk '{print $5}'

or

ls -l somefile | awk '{print $11}'

Perfect!

That's exactly what I needed.

I suppose I could have read the 'awk' man page but at the time it seemed so daunting, at least now I understand the basics of it.

Thank you!

P.S. There you go Jeremy (and macemoneta) yet another satified Linux user, at least untill I come up with another Question! :)

Well after a long time of not thinking about this I was poking around and found the readlink command:

The awk method does work and for more that just the question I had. But for finding the target of a symlink I think this works better. (Much simpler, easyer to read, remember what it does in a script, etc.)

Also it is contained in the coreutils package (Ubuntu) so most if not all distros will have it.

One more reason to use readlink is that it can follow recursively
 

I know is an extremely old thread, but hey, some of these tips are still valid after years!!!

I just wanted to point out one advantage of using readlink instead of the awk solution is that readlink can follow links recursively as opposed to the awk solution.
for example, I found this thread trying to find a way to get the real java command in my system (in a script), so I ended up doing this:

Which in fact has two symlinks: