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

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.

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"

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.

If you want it without the path then you would use ${0##*/}

To answer Chris Conway, on Linux (at least) you would do this:

echo $(basename $(readlink -nf $0))

readlink prints out the value of a symbolic link. If it isn't a symbolic link, it prints the file name. -n tells it to not print a newline. -f tells it to follow the link completely (if a symbolic link was a link to another link, it would resolve that one as well).

Since some comments asked about the filename without extension, here's an example how to accomplish that:

FileName=${0##*/}
FileNameWithoutExtension=${FileName%.*}

Enjoy!

These answers are correct for the cases they state but there is a still a problem if you run the script from another script using the 'source' keyword (so that it runs in the same shell). In this case, you get the $0 of the calling script. And in this case, I don't think it is possible to get the name of the script itself.

This is an edge case and should not be taken TOO seriously. If you run the script from another script directly (without 'source'), using $0 will work.

You can use $0 to determine your script name (with full path) - to get the script name only you can trim that variable with

basename $0

This works fine with ./self.sh, ~/self.sh, source self.sh, source ~/self.sh:

#!/usr/bin/env bash

self=$(readlink -f "${BASH_SOURCE[0]}")
basename=$(basename "$self")

echo "$self"
echo "$basename"

Credits: I combined multiple answers to get this one.

In bash you can get the script file name using $0. Generally $1, $2 etc are to access CLI arguments. Similarly $0 is to access the name which triggers the script(script file name).

#!/bin/bash
echo "You are running $0"
...
...

If you invoke the script with path like /path/to/script.sh then $0 also will give the filename with path. In that case need to use $(basename $0) to get only script file name.

Short, clear and simple, in my_script.sh

#!/bin/bash

running_file_name=$(basename "$0")

echo "You are running '$running_file_name' file."

Out put:

./my_script.sh
You are running 'my_script.sh' file.

Here is what I came up with, inspired by Dimitre Radoulov's answer (which I upvoted, by the way).

script="$BASH_SOURCE"
[ -z "$BASH_SOURCE" ] && script="$0"

echo "Called $script with $# argument(s)"

regardless of the way you call your script

. path/to/script.sh

or

./path/to/script.sh

$0 will give the name of the script you are running. Create a script file and add following code

#!/bin/bash
echo "Name of the file is $0"

then run from terminal like this

./file_name.sh

echo "You are running $0"