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 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.

# ------------- 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 ------------- #

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 '$0 is: %sn$BASH_SOURCE is: %sn' "$0" "$BASH_SOURCE"

$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) 

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).

I’ve found this line to always work, regardless of whether the file is being sourced or run as a script.

echo "${BASH_SOURCE[${#BASH_SOURCE[@]} - 1]}"

If you want to follow symlinks use readlink on the path you get above, recursively or non-recursively.

The reason the one-liner works is explained by the use of the BASH_SOURCE environment variable and its associate FUNCNAME.

BASH_SOURCE

An array variable whose members are the source filenames where the corresponding shell function names in the FUNCNAME array variable are defined. The shell function ${FUNCNAME[$i]} is defined in the file ${BASH_SOURCE[$i]} and called from ${BASH_SOURCE[$i+1]}.

FUNCNAME

An array variable containing the names of all shell functions currently in the execution call stack. The element with index 0 is the name of any currently-executing shell function. The bottom-most element (the one with the highest index) is “main”. This variable exists only when a shell function is executing. Assignments to FUNCNAME have no effect and return an error status. If FUNCNAME is unset, it loses its special properties, even if it is subsequently reset.

This variable can be used with BASH_LINENO and BASH_SOURCE. Each element of FUNCNAME has corresponding elements in BASH_LINENO and BASH_SOURCE to describe the call stack. For instance, ${FUNCNAME[$i]} was called from the file ${BASH_SOURCE[$i+1]} at line number ${BASH_LINENO[$i]}. The caller builtin displays the current call stack using this information.

[Source: Bash manual]

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.

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

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

Enjoy!

Re: Tanktalus’s (accepted) answer above, a slightly cleaner way is to use:

me=$(readlink --canonicalize --no-newline $0)

If your script has been sourced from another bash script, you can use:

me=$(readlink --canonicalize --no-newline $BASH_SOURCE)

I agree that it would be confusing to dereference symlinks if your objective is to provide feedback to the user, but there are occasions when you do need to get the canonical name to a script or other file, and this is the best way, imo.

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="$(dirname "$(realpath "$BASH_SOURCE")")"

This resolves symbolic links (realpath does that), handles spaces (double quotes do this), and will find the current script name even when sourced (. ./myscript) or called by other scripts ($BASH_SOURCE handles that). After all that, it is good to save this in a environment variable for re-use or for easy copy elsewhere (this=)…

if your invoke shell script like

/home/mike/runme.sh

$0 is full name

 /home/mike/runme.sh

basename $0 will get the base file name

 runme.sh

and you need to put this basic name into a variable like

filename=$(basename $0)

and add your additional text

echo "You are running $filename"

so your scripts like

/home/mike/runme.sh
#!/bin/bash 
filename=$(basename $0)
echo "You are running $filename"

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.

echo "$(basename "`test -L ${BASH_SOURCE[0]} 
                   && readlink ${BASH_SOURCE[0]} 
                   || echo ${BASH_SOURCE[0]}`")"

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.

Info thanks to Bill Hernandez. I added some preferences I’m adopting.

#!/bin/bash
function Usage(){
    echo " Usage: show_parameters [ arg1 ][ arg2 ]"
}
[[ ${#2} -eq 0 ]] && Usage || {
    echo
    echo "# arguments called with ---->  ${@}     "
    echo "# $1 ----------------------->  $1       "
    echo "# $2 ----------------------->  $2       "
    echo "# path to me --------------->  ${0}     " | sed "s/$USER/$USER/g"
    echo "# parent path -------------->  ${0%/*}  " | sed "s/$USER/$USER/g"
    echo "# my name ------------------>  ${0##*/} "
    echo
}

Cheers

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

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.

echo “You are running $0”

DIRECTORY=$(cd `dirname $0` && pwd)

I got the above from another Stack¬†Overflow question, Can a Bash script tell what directory it’s stored in?, but I think it’s useful for this topic as well.

$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

somthing like this?

export LC_ALL=en_US.UTF-8
#!/bin/bash
#!/bin/sh

#----------------------------------------------------------------------
start_trash(){
ver="htrash.sh v0.0.4"
$TRASH_DIR  # url to trash $MY_USER
$TRASH_SIZE # Show Trash Folder Size

echo "Would you like to empty Trash  [y/n]?"
read ans
if [ $ans = y -o $ans = Y -o $ans = yes -o $ans = Yes -o $ans = YES ]
then
echo "'yes'"
cd $TRASH_DIR && $EMPTY_TRASH
fi
if [ $ans = n -o $ans = N -o $ans = no -o $ans = No -o $ans = NO ]
then
echo "'no'"
fi
 return $TRUE
} 
#-----------------------------------------------------------------------

start_help(){
echo "HELP COMMANDS-----------------------------"
echo "htest www                 open a homepage "
echo "htest trash               empty trash     "
 return $TRUE
} #end Help
#-----------------------------------------------#

homepage=""

return $TRUE
} #end cpdebtemp

# -Case start
# if no command line arg given
# set val to Unknown
if [ -z $1 ]
then
  val="*** Unknown  ***"
elif [ -n $1 ]
then
# otherwise make first arg as val
  val=$1
fi
# use case statement to make decision for rental
case $val in
   "trash") start_trash ;;
   "help") start_help ;;
   "www") firefox $homepage ;;
   *) echo "Sorry, I can not get a $val   for you!";;
esac
# Case stop

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