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?
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).
# ------------- 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 ------------- #
随着bash> = 3,以下工作:
$ ./s
0 is: ./s
BASH_SOURCE is: ./s
$ . ./s
0 is: bash
BASH_SOURCE is: ./s
$ cat s
#!/bin/bash
printf '$0 is: %sn$BASH_SOURCE is: %sn' "$0" "$BASH_SOURCE"
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