How do I know the script file name in a Bash script?
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"
# ------------- 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 ------------- #
$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)
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
.