mjhjmtu

In the most common cases, $0 will contain a path, absolute or relative to the script, so

script_path=$(readlink -e -- "$0")

(assuming there's a readlink command and it supports -e) generally is a good enough way to obtain the canonical absolute path to the script.

$0 is assigned from the argument specifying the script as passed to the interpreter.

For example, in:

the-shell -shell-options the/script its args

$0 gets the/script.

When you run:

the/script its args

Your shell will do a:

exec("the/script", ["the/script", "its", "args"])

If the script contains a #! /bin/sh - she-bang for instance, the system will transform that to:

exec("/bin/sh", ["/bin/sh" or "the/script", "-", "the/script", "its", "args"])

(if it doesn't contain a she-bang, or more generally if the system returns a ENOEXEC error, then its your shell that will do the same thing)

There's an exception for setuid/setgid scripts on some systems, where the system will open the script on some fd x and run instead:

exec("/bin/sh", ["/bin/sh" or "the/script", "-", "/dev/fd/x", "its", "args"])

to avoid race conditions (in which case $0 will contain /dev/fd/x).

Now, you may argue that /dev/fd/x is a path to that script. Note however that if you read from $0, you'll break the script as you consume the input.

Now, there's a difference if the script command name as invoked doesn't contain a slash. In:

the-script its args

Your shell will look up the-script in $PATH. $PATH may contain absolute or relative (including the empty string) paths to some directories. For instance, if $PATH contains /bin:/usr/bin: and the-script is found in the current directory, the shell will do a:

exec("the-script", ["the-script", "its", "args"])

which will become:

exec("/bin/sh", ["/bin/sh" or "the-script", "-", "the-script", "its", "args"]

Or if it's found in /usr/bin:

exec("/usr/bin/the-script", ["the-script", "its", "args"])
exec("/bin/sh", ["/bin/sh" or "the-script" or "/usr/bin/the-script",
 "-", "/usr/bin/the-script", "its", "args")

In all those cases above except the setuid corner case, $0 will contain a path (absolute or relative) to the script.

Now, a script can also be called as:

the-interpreter the-script its args

When the-script as above doesn't contain slash characters, the behaviour varies slightly from shell to shell.

Old AT&T ksh implementations were actually looking up the-script unconditionally in $PATH (which was actually a bug and a security hole for setuid scripts), so $0 actually did not contain a path to the script unless the $PATH lookup actually happened to find the-script in the current directory.

Newer AT&T ksh would try and interpret the-script in the current directory if it's readable. If not it would lookup for a readable and executable the-script in $PATH.

For bash, it checks if the-script is in the current directory (and is not a broken symlink) and if not, lookup for a readable (not necessarily executable) the-script in $PATH.

zsh in sh emulation would do like bash except that if the-script is a broken symlink in the current directory, it would not search for a the-script in $PATH and would instead report an error.

All the other Bourne-like shells don't look the-script up in $PATH.

For all those shells anyway, if you find that $0 doesn't contain a / and is not readable, then it probably has been looked up in $PATH. Then, as files in $PATH are likely to be executable, it's probably a safe approximation to use command -v -- "$0" to find its path (though that wouldn't work if $0 happens to also be the name of a shell builtin or keyword (in most shells)).

So if you really want to cover for that case, you could write it:

progname=$0
[ -r "$progname" ] || progname=$(
 IFS=:; set -f
 for i in $PATH-$(getconf PATH)""; do
 case $i in
 "") p=$progname;;
 */) p=$i$progname;;
 *) p=$i/$progname
 esac
 [ -r "$p" ] && exec printf '%sn' "$p"
 done
 exit 1
 ) && progname=$(readlink -e -- "$progname") ||
 progname=unknown

(the "" appended to $PATH is to preserve a trailing empty element with shells whose $IFS acts as delimiter instead of separator).

Now, there are more esoteric ways to invoke a script. One could do:

the-shell < the-script

Or:

cat the-script | the-shell

In that case, $0 will be the first argument (argv[0]) that the interpreter received (above the-shell, but that could be anything though generally either the basename or one path to that interpreter).

Detecting that you're in that situation based on the value of $0 is not reliable. You could look at the output of ps -o args= -p "$$" to get a clue. In the pipe case, there's no real way you can get back to a path to the script.

One could also do:

the-shell -c '. the-script' blah blih

Then, except in zsh (and some old implementation of the Bourne shell), $0 would be blah. Again, hard to get to the path of the script in those shells.

Or:

the-shell -c "$(cat the-script)" blah blih

etc.

To make sure you have the right $progname, you could search for a specific string in it like:

progname=$0
[ -r "$progname" ] || progname=$(
 IFS=:; set -f
 for i in $PATH-$(getconf PATH):; do
 case $i in
 "") p=$progname;;
 */) p=$i$progname;;
 *) p=$i/$progname
 esac
 [ -r "$p" ] && exec printf '%sn' "$p"
 done
 exit 1
 ) && progname=$(readlink -e -- "$progname") ||
 progname=unknown

[ -f "$progname" ] && grep -q 7YQLVVD3UIUDTA32LSE8U9UOHH < "$progname" ||
 progname=unknown

But again I don't think it's worth the effort.

Here are two situations where the directory would not be included:

> bash scriptname
scriptname

> bash <scriptname
bash

In both cases, the current directory would have to be the directory where scriptname was located.

In the first case, the value of $0 could still be passed to grep since it assumes the FILE argument is relative to the current directory.

In the second case, if the help and version information is only printed in response to a specific command line option, it shouldn't be a problem. I'm not sure why anyone would invoke a script that way to print the help or version information.

Caveats

• If the script changes the current directory, you wouldn't want to use relative paths.




• If the script is sourced, the value of $0 will usually be the caller script rather than the sourced script.

An arbitrary argument zero can be specified when using the -c option to most (all?) shells. Eg:

sh -c 'echo $0' argv0

From man bash (chosen purely because this has a better description than my man sh - usage is the same regardless):

-c

If the -c option is present, then commands are read from the first non-option argument command_string. If there are arguments after the command_string, they are
assigned to the positional parameters, starting with $0.

NOTE: Other's have already explained the mechanics of $0 so I'll skip all that.

I generally side step this entire issue and simply use the command readlink -f $0. This will always give you back the full path of whatever you give to it as an argument.

Examples

Say I'm here to start with:

$ pwd
/home/saml/tst/119929/adir

Make a directory + file:

$ mkdir adir
$ touch afile
$ cd adir/

Now start showing off readlink:

$ readlink -f ../adir
/home/saml/tst/119929/adir

$ readlink -f ../
/home/saml/tst/119929

$ readlink -f ../afile 
/home/saml/tst/119929/afile

$ readlink -f .
/home/saml/tst/119929/adir

Additional trickery

Now with a consistent result being returned when we interrogate $0 via readlink we can use a simply dirname $(readlink -f $0) to get the absolute path to the script -or- basename $(readlink -f $0) to get the actual name of the script.

My man page says:

$0 : expands to the name of shell or shell script.

It seems this translates to argv[0] of the current shell - or the first nonoperand commandline argument the currently interpreting shell is fed when invoked. I previously stated that sh ./somescript would path its $0 $ENV variable to sh but this was incorrect because the shell is a new process of its own and invoked with a new $ENV.

In this way sh ./somescript.sh differs from . ./somescript.sh which runs in the current environment and $0 is already set.

You can check this by comparing $0 to /proc/$$/status.

echo 'script="/proc/$$/status"
 echo $0
 cat "$script"' 
 > ./script.sh
sh ./script.sh ; . ./script.sh

Thanks for the correction, @toxalot. I learned something.

I want to grep the current script so I can print out help and version information from the comments section at the top.

While $0 contains the script name it may contain a prefixed path based on the way the script is called, I have always used $0##*/ to print the script name in the help output which removes any leading path from $0.

Taken from Advanced Bash Scripting guide - Section 10.2 parameter substitution

$var#Pattern

Remove from $var the shortest part of $Pattern that matches the front end of $var.

$var##Pattern

Remove from $var the longest part of $Pattern that matches the front end of $var.

So the longest part of $0 that matches */ will be the entire path prefix, returning only the script name.

For something similar I use:

rPath="$(dirname $(realpath $0))"
echo $rPath 

rPath=$(dirname $(readlink -e -- "$0"))
echo $rPath 

rPath always has the same value.