In the bash script how do I know the script file name?

Question

How can I determine the name of the bash script file inside the script itself?
Like if my script is in file runme.sh, than how would I make it to display "You are running runme.sh" message without hardcodding that?

Thanks,

Answer 1

me=`basename $0`

For reading through a symlink, which is usually not what you want (you usually don't want to confuse the user this way), try:

me="$(basename "$(test -L "$0" && readlink "$0" || echo "$0")")"

IMO, that'll produce confusing output. "I ran foo.sh, but it's saying I'm running bar.sh!? Must be a bug!" Besides, one of the purposes of having differently-named symlinks is to provide different functionality based on the name it's called as (think gzip and gunzip on some platforms).

Answer 2

echo "You are running $0"

Answer 3

You can use $0 to determine your script name (with full path) - to get the script name only you can trim that variable with

basename $0

Answer 4

If the script name has spaces in it, a more robust way is to use "$0" or "$(basename "$0")" to prevent the name from getting mangled or interpreted in any way. In general, it is good practice to always quote variable names in the shell.

Answer 5

$0 doesn't answer the question (as I understand it). A demonstration:

$ cat script.sh
#! /bin/sh
echo `basename $0`
$ ./script.sh 
script.sh
$ ln script.sh linktoscript
$ ./linktoscript 
linktoscript

How does one get ./linktoscript to print out script.sh?

[EDIT] Per @ephemient in comments above, though the symbolic link thing may seem contrived, it is possible to fiddle with $0 such that it does not represent a filesystem resource. The OP is a bit ambiguous about what he wanted.

Answer 6

To answer Chris Conway, on Linux (at least) you would do this:

echo $(basename $(readlink -nf $0))

readlink prints out the value of a symbolic link. If it isn't a symbolic link, it prints the file name. -n tells it to not print a newline. -f tells it to follow the link completely (if a symbolic link was a link to another link, it would resolve that one as well).

Answer 7

If you want it without the path then you would use ${0##*/}

Answer 8

These answers are correct for the cases they state but there is a still a problem if you run the script from another script using the 'source' keyword (so that it runs in the same shell). In this case, you get the $0 of the calling script. And in this case, I don't think it is possible to get the name of the script itself.

This is an edge case and should not be taken TOO seriously. If you run the script from another script directly (without 'source'), using $0 will work.

Answer 9

With bash >= 3:

$ ./s
$0 is: ./s
$BASH_SOURCE is: ./s
$ . ./s
$0 is: bash
$BASH_SOURCE is: ./s

$ cat s
#!/bin/bash

printf '$0 is: %s\n$BASH_SOURCE is: %s\n' "$0" "$BASH_SOURCE

Answer 10

# ------------- SCRIPT ------------- #

#!/bin/bash

echo
echo "# arguments called with ---->  ${@}     "
echo "# \$1 ---------------------->  $1       "
echo "# \$2 ---------------------->  $2       "
echo "# path to me --------------->  ${0}     "
echo "# parent path -------------->  ${0%/*}  "
echo "# my name ------------------>  ${0##*/} "
echo
exit

# ------------- CALLED ------------- #

# Notice on the next line, the first argument is called within double, 
# and single quotes, since it contains two words

$  /misc/shell_scripts/check_root/show_parms.sh "'hello there'" "'william'"

# ------------- RESULTS ------------- #

# arguments called with --->  'hello there' 'william'
# $1 ---------------------->  'hello there'
# $2 ---------------------->  'william'
# path to me -------------->  /misc/shell_scripts/check_root/show_parms.sh
# parent path ------------->  /misc/shell_scripts/check_root
# my name ----------------->  show_parms.sh

# ------------- END ------------- #

Answer 11

Re: Tanktalus's (accepted) answer above, a slightly cleaner way is to use:

me=$(readlink --canonicalize --no-newline $0)

If your script has been sourced from another bash script, you can use:

me=$(readlink --canonicalize --no-newline $BASH_SOURCE)

I agree that it would be confusing to dereference symlinks if your objective is to provide feedback to the user, but there are occasions when you do need to get the canonical name to a script or other file, and this is the best way, imo.