Question

We have scripts of following nature (in cron)

someScript.sh > /tmp/cronlog/somescript.$(date +%Y%m%d).log 2>&1

Now is there a way by which with in someScript.sh I can figure out what file the output has gone in to?

The script sends email with summary. At the same time I would like to mention that details could be found in so and so output file – with in the email.

I am aware of the construct if [ -t 1 ] to detect stdout etc but how to get the output file name?
Note that I want this to be generic so that some one can change the output file in cron and the script does not need to be modified.

Answer 1

The simplest thing I could think is that:

readlink -f /proc/$$/fd/1

$$ is the PID of the script (inside the script). On most unix systems, /proc/[pid] is the pseudo-directory containing info for process [pid].

/proc/[pid]/fd is a directory containing a list of symlinks for the open file-descriptors of the process. fd/0 is input, fd/1 is the output of the script, etc.

readlink then gives you the target file or tty if you don’t redirect the output.

Of course, if you want to display it, you have to display it somewhere else than standard ouput, or it will be redirected! To debug, try the std error (2).

Various callings give those results on my box (script.sh just calls readlink -f /proc/$$/fd/1 >&2)

# ./script.sh
/dev/pts/0

# ./script.sh > /var/tmp/foo
/var/tmp/foo

# ./script.sh | more
/proc/12132/fd/pipe:[916212]