How do I know the script file name in a Bash script?

How can I determine the name of the Bash script file inside the script itself?

Like if my script is in file runme.sh, then how would I make it to display "You are running runme.sh" message without hardcoding that?


Solution 1:

me=`basename "$0"`

For reading through a symlink1, 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).


1 That is, to resolve symlinks such that when the user executes foo.sh which is actually a symlink to bar.sh, you wish to use the resolved name bar.sh rather than foo.sh.

Solution 2:

# ------------- 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 ------------- #

Solution 3:

With bash >= 3 the following works:

$ ./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"

Solution 4:

$BASH_SOURCE gives the correct answer when sourcing the script.

This however includes the path so to get the scripts filename only, use:

$(basename $BASH_SOURCE) 

Solution 5:

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