I have the following code in my bash script. Now I wanna use it in POSIX sh. How can I convert it?
DIR="$( cd "$( dirname "${BASH_SOURCE[0]}" )" > /dev/null && pwd )"
The POSIX-shell (sh
) counterpart of $BASH_SOURCE
is $0
. see bottom for background info
Caveat: The crucial difference is that if your script is being sourced (loaded into the current shell with .
), the snippets below will not work properly. explanation further below
Note that I've changed DIR
to dir
in the snippets below, because it's better not to use all-uppercase variable names so as to avoid clashes with environment variables and special shell variables.
The CDPATH=
prefix takes the place of > /dev/null
in the original command: $CDPATH
is set to a null string so as to ensure that cd
never echoes anything.
In the simplest case, this will do (the equivalent of the OP's command):
dir=$(CDPATH= cd -- "$(dirname -- "$0")" && pwd)
If you also want to resolve the resulting directory path to its ultimate target in case the directory and/or its components are symlinks, add -P
to the pwd
command:
dir=$(CDPATH= cd -- "$(dirname -- "$0")" && pwd -P)
Caveat: This is NOT the same as finding the script's own true directory of origin:
Let's say your script foo
is symlinked to /usr/local/bin/foo
in the $PATH
, but its true path is /foodir/bin/foo
.
The above will still report /usr/local/bin
, because the symlink resolution (-P
) is applied to the directory, /usr/local/bin
, rather than to the script itself.
To find the script's own true directory of origin, you'd have to inspect the script's path to see if it's a symlink and, if so, follow the (chain of) symlinks to the ultimate target file, and then extract the directory path from the target file's canonical path.
GNU's readlink -f
(better: readlink -e
) could do that for you, but readlink
is not a POSIX utility.
While BSD platforms, including macOS, have a readlink
utility too, on macOS it doesn't support -f
's functionality. That said, to show how simple the task becomes if readlink -f
is available: dir=$(dirname "$(readlink -f -- "$0")")
.
In fact, there is no POSIX utility for resolving file symlinks. There are ways to work around that, but they're cumbersome and not fully robust:
The following, POSIX-compliant shell function implements what GNU's readlink -e
does and is a reasonably robust solution that only fails in two rare edge cases:
->
(also rare)With this function, named rreadlink
, defined, the following determines the script's true directory path of origin:
dir=$(dirname -- "$(rreadlink "$0")")
Note: If you're willing to assume the presence of a (non-POSIX) readlink
utility - which would cover macOS, FreeBSD and Linux - a similar, but simpler solution can be found in this answer to a related question.
rreadlink()
source code - place before calls to it in scripts:
rreadlink() ( # Execute the function in a *subshell* to localize variables and the effect of `cd`.
target=$1 fname= targetDir= CDPATH=
# Try to make the execution environment as predictable as possible:
# All commands below are invoked via `command`, so we must make sure that `command`
# itself is not redefined as an alias or shell function.
# (Note that command is too inconsistent across shells, so we don't use it.)
# `command` is a *builtin* in bash, dash, ksh, zsh, and some platforms do not even have
# an external utility version of it (e.g, Ubuntu).
# `command` bypasses aliases and shell functions and also finds builtins
# in bash, dash, and ksh. In zsh, option POSIX_BUILTINS must be turned on for that
# to happen.
{ \unalias command; \unset -f command; } >/dev/null 2>&1
[ -n "$ZSH_VERSION" ] && options[POSIX_BUILTINS]=on # make zsh find *builtins* with `command` too.
while :; do # Resolve potential symlinks until the ultimate target is found.
[ -L "$target" ] || [ -e "$target" ] || { command printf '%s\n' "ERROR: '$target' does not exist." >&2; return 1; }
command cd "$(command dirname -- "$target")" # Change to target dir; necessary for correct resolution of target path.
fname=$(command basename -- "$target") # Extract filename.
[ "$fname" = '/' ] && fname='' # !! curiously, `basename /` returns '/'
if [ -L "$fname" ]; then
# Extract [next] target path, which may be defined
# *relative* to the symlink's own directory.
# Note: We parse `ls -l` output to find the symlink target
# which is the only POSIX-compliant, albeit somewhat fragile, way.
target=$(command ls -l "$fname")
target=${target#* -> }
continue # Resolve [next] symlink target.
fi
break # Ultimate target reached.
done
targetDir=$(command pwd -P) # Get canonical dir. path
# Output the ultimate target's canonical path.
# Note that we manually resolve paths ending in /. and /.. to make sure we have a normalized path.
if [ "$fname" = '.' ]; then
command printf '%s\n' "${targetDir%/}"
elif [ "$fname" = '..' ]; then
# Caveat: something like /var/.. will resolve to /private (assuming /var@ -> /private/var), i.e. the '..' is applied
# AFTER canonicalization.
command printf '%s\n' "$(command dirname -- "${targetDir}")"
else
command printf '%s\n' "${targetDir%/}/$fname"
fi
)
To be robust and predictable, the function uses command
to ensure that only shell builtins or external utilities are called (ignores overloads in the forms of aliases and functions).
It's been tested in recent versions of the following shells: bash
, dash
, ksh
, zsh
.
tl;dr:
Using POSIX features only:
zsh
, which, however, doesn't usually act as sh
).$0
to the shell executable name/path (except in zsh
, where, as noted $0
is truly the current script's path). By contrast (except in zsh
), a script being sourced from another script that itself was directly invoked, contains that script's path in $0
.bash
, ksh
, and zsh
have nonstandard features that do allow determining the actual script path even in sourced scenarios and also detecting whether a script is being sourced or not; for instance, in bash
, $BASH_SOURCE
always contains the running script's path, whether it's being sourced or not, and [[ $0 != "$BASH_SOURCE" ]]
can be used to test whether the script is being sourced.To show why this cannot be done, let's analyze the command from Walter A's answer:
# NOT recommended - see discussion below.
DIR=$( cd -P -- "$(dirname -- "$(command -v -- "$0")")" && pwd -P )
-P
twice is redundant - it's sufficient to use it with pwd
.cd
's potential stdout output, if $CDPATH
happens to be set.)command -v -- "$0"
command -v -- "$0"
is designed to cover one additional scenario: if the script is being sourced from an interactive shell, $0
typically contains the mere filename of the shell executable (sh
), in which case dirname
would simply return .
(because that's what dirname
invariably does when given a argument without a path component).
command -v -- "$0"
then returns that shell's absolute path through a $PATH
lookup (/bin/sh
). Note, however, that login shells on some platforms (e.g., OSX) have their filename prefixed with -
in $0
(-sh
), in which case command -v -- "$0"
doesn't work as intended (returns an empty string).command -v -- "$0"
can misbehave in two non-sourced scenarios in which the shell executable, sh
, is directly invoked, with the script as an argument:
command -v -- "$0"
may return an empty string, depending on what specific shell acts as sh
on a given system: bash
, ksh
, and zsh
return an empty string; only dash
echoes $0
command
doesn't explicitly say whether command -v
, when applied to a filesystem path, should only return executable files - which is what bash
, ksh
, and zsh
do - but you can argue that it is implied by the very purpose of command
; curiously, dash
, which is usually the most compliant POSIX citizen, is deviating from the standard here. By contrast, ksh
is the lone model citizen here, because it is the only one that reports executable files only and reports them with an absolute (albeit not normalized) path, as the spec requires.$PATH
, and the invocation uses its mere filename (e.g., sh myScript
), command -v -- "$0"
will also return the empty string, except in dash
.$0
then doesn't contain that information (except in zsh
, which doesn't usually act as sh
) - there's no good solution to this problem.
$0
directly: [ "$0" = "sh" ] || [ "$0" = "-sh" ] || [ "$0" = "/bin/sh" ]
$0
then simply contains the sourcing script's path.command -v -- "$0"
in sourced scenarios and the fact that it breaks two non-sourced scenarios, my vote is for NOT using it, which leaves us with:
$dir
ends up either containing .
, if the shell executable was invoked as a mere filename (applying dirname
to a mere filename always returns .
), or the shell executable's directory path otherwise. .
cannot be reliably distinguished from a non-sourced invocation from the current directory.$0
contains that script's path, and the script being sourced has no way of telling whether that's the case.POSIX defines the behavior of $0
with respect to shell scripts here.
Essentially, $0
should reflect the path of the script file as specified, which implies:
$0
containing an absolute path.$0
contains an absolute path only if:
~/bin/myScript
(assuming the script itself is executable)sh ~/bin/myScript
$PATH
; behind the scenes, the system transforms myScript
into an absolute path and then executes it; e.g.:
myScript # executes /home/jdoe/bin/myScript, for instance
In all other cases, $0
will reflect the script path as specified:
sh
with a script, this can be a mere filename (e.g., sh myScript
) or a relative path (e.g., sh ./myScript
)./myScript
- note that a mere filename would only find scripts in the $PATH
).In practice, bash
, dash
, ksh
, and zsh
all exhibit this behavior.
By contrast, POSIX does NOT mandate the value of $0
when sourcing a script (using the special built-in utility .
("dot")), so you cannot rely on it, and, in practice, behavior differs across shells.
$0
when your script is being sourced and expect standardized behavior.
bash
, dash
, and ksh
leave $0
untouched when sourcing scripts, meaning that $0
contains the caller's $0
value, or, more accurately, the $0
value of the most recent caller in the call chain that hasn't been sourced itself; thus, $0
may point either to the shell's executable or to the path of another (directly invoked) script that sourced the current one. zsh
, as the lone dissenter, actually does report the current script's path in $0
. Conversely, $0
will provide no indication as to whether the script is being sourced or not.$0
to the current script's path is.bash
, ksh
, and zsh
all offer their own ways of obtaining the running script's path, even when it's being sourced.For the sake of completeness: the value of $0
in other contexts:
$0
remain unchanged; therefore, whatever value it has outside the function, it'll have inside as well.
bash
, dash
, and ksh
do behave that way.zsh
is the lone dissenter and reports the function's name.-c
option on startup, it's the first operand (non-option argument) that sets $0
; e.g.:
sh -c 'echo \$0: $0 \$1: $1' foo one # -> '$0: foo $1: one'
bash
, dash
, ksh
, and zsh
all behave that way.$0
is the value of the first argument that the shell's parent process passed - typically, that's the shell's name or path (e.g. sh
, or /bin/sh
); this includes:
-
to the shell name before placing it in $0
, so as to signal to the shell that it is a _login shell; thus, by default, $0
reports -bash
, not bash
, in interactive shells on OSX.sh < myScript
)bash
, dash
, ksh
, and zsh
all behave that way.