Question

I want to write a very simple script , which takes a process name , and return the tail of the last file name which contains the process name.

I wrote something like that :

#!/bin/sh
tail $(ls -t *"$1"*| head -1) -f

My question:

1. Do I need the first line?

2. Why isn’t ls -t *"$1"*| head -1 | tail -f working?

3. Is there a better way to do it?

Answer 1

1: The first line is a so called she-bang, read the description here:

In computing, a shebang (also called a
hashbang, hashpling, pound bang, or
crunchbang) refers to the characters
“#!” when they are the first two
characters in an interpreter directive
as the first line of a text file. In a
Unix-like operating system, the
program loader takes the presence of
these two characters as an indication
that the file is a script, and tries
to execute that script using the
interpreter specified by the rest of
the first line in the file

2: tail can’t take the filename from the stdin: It can either take the text on the stdin or a file as parameter. See the man page for this.

3: No better solution comes to my mind: Pay attention to filenames containing spaces: This does not work with your current solution, you need to add quotes around the $() block.