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

 # ------------- 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 '$0 is: %s\n$BASH_SOURCE is: %s\n' "$0" "$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]

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.

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.

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.

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=)...

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 

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" 

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 "$(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

To get the "realpath" of script or sourced scripts in all cases :

fullname=$(readlink $0) # Take care of symbolic links dirname=${fullname%/*} # Get (most of the time) the dirname realpath=$(dirname $BASH_SOURCE) # TO handle sourced scripts [ "$realpath" = '.' ] && realpath=${dirname:-.} 

Here is the bash script to generate (in a newly created "workdir" subdir and in "mytest" in current dir), a bash script which in turn will source another script, which in turm will call a bash defined function .... tested with many ways to launch them :

#!/bin/bash ############################################################## ret=0 fullname=$(readlink $0) # Take care of symbolic links dirname=${fullname%/*} # Get (most of the time) the dirname realpath=$(dirname $BASH_SOURCE) # TO handle sourced scripts [ "$realpath" = '.' ] && realpath=${dirname:-.} fullname_withoutextension=${fullname%.*} mkdir -p workdir cat <<'EOD' > workdir/_script_.sh #!/bin/bash ############################################################## ret=0 fullname=$(readlink $0) # Take care of symbolic links dirname=${fullname%/*} # Get (most of the time) the dirname realpath=$(dirname $BASH_SOURCE) # TO handle sourced scripts [ "$realpath" = '.' ] && realpath=${dirname:-.} fullname_withoutextension=${fullname%.*} echo echo "# ------------- RESULTS ------------- #" echo "# path to me (\$0)-----------> ${0} " echo "# arguments called with ----> ${@} " echo "# \$1 -----------------------> $1 " echo "# \$2 -----------------------> $2 " echo "# path to me (\$fullname)----> ${fullname} " echo "# parent path(\${0%/*})------> ${0%/*} " echo "# parent path(\$dirname)-----> ${dirname} " echo "# my name ----\${0##*/}------> ${0##*/} " echo "# my source -\${BASH_SOURCE}-> ${BASH_SOURCE} " echo "# parent path(from BASH_SOURCE) -> $(dirname $BASH_SOURCE)" echo "# my function name -\${FUNCNAME[0]}------> ${FUNCNAME[0]}" echo "# my source or script real path (realpath)------------------> $realpath" echo [ "$realpath" = "workdir" ] || ret=1 [ $ret = 0 ] || echo "*******************************************************" [ $ret = 0 ] || echo "*********** ERROR **********************************" [ $ret = 0 ] || echo "*******************************************************" show_params () { echo echo "# --- RESULTS FROM show_params() ---- #" echo "# path to me (\$0)-----------> ${0} " echo "# arguments called with ----> ${@} " echo "# \$1 -----------------------> $1 " echo "# \$2 -----------------------> $2 " echo "# path to me (\$fullname)----> ${fullname} " echo "# parent path(\${0%/*})------> ${0%/*} " echo "# parent path(\$dirname)-----> ${dirname} " echo "# my name ----\${0##*/}------> ${0##*/} " echo "# my source -\${BASH_SOURCE}-> ${BASH_SOURCE} " echo "# parent path(from BASH_SOURCE) -> $(dirname $BASH_SOURCE)" echo "# my function name -\${FUNCNAME[0]}------> ${FUNCNAME[0]}" echo "# my source or script real path (realpath)------------------> $realpath" echo [ "$realpath" = "workdir" ] || ret=1 [ $ret = 0 ] || echo "*******************************************************" [ $ret = 0 ] || echo "*********** ERROR **********************************" [ $ret = 0 ] || echo "*******************************************************" } show_params "$@" EOD cat workdir/_script_.sh > workdir/_side_by_side_script_sourced.inc cat <<'EOD' >> workdir/_script_.sh echo "# . $realpath/_side_by_side_script_sourced.inc 'hello there' 'william'" . $realpath/_side_by_side_script_sourced.inc 'hello there' 'william' [ $ret = 0 ] || echo "*******************************************************" [ $ret = 0 ] || echo "*********** ERROR **********************************" [ $ret = 0 ] || echo "*******************************************************" EOD chmod +x workdir/_script_.sh [ -L _mytest_ ] && rm _mytest_ ln -s workdir/_script_.sh _mytest_ # ------------- CALLED ------------- # called_by () { echo '==========================================================================' echo " Called by : " "$@" echo '==========================================================================' eval "$@" } called_by bash _mytest_ called_by ./_mytest_ called_by bash workdir/_script_.sh called_by workdir/_script_.sh called_by . workdir/_script_.sh # ------------- RESULTS ------------- # echo echo [ $ret = 0 ] || echo "*******************************************************" [ $ret = 0 ] || echo "*********** ERROR **********************************" [ $ret = 0 ] || echo "*******************************************************" echo [ $ret = 0 ] && echo ".... location of scripts (\$realpath) should always be equal to $realpath, for all test cases at date". echo # ------------- END ------------- # 

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.

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 

As an option not listed in above answers you can use script's pid to get a lot of info including name from /proc/$pid folder, like this:

#!/bin/bash pid=$$ # get script's pid to var $pid echo $pid # there is a file in /proc/$pid folder called cmdline # this file stores full command line used by script cat /proc/$pid/cmdline 

output:

$ ./test 1 '2 3' 4 12924 /bin/bash./test12 34 

the output looks a bit messy it's just one string... not exactly it actually has delimiters but to be able to see them we need to change our cat command like this:

#!/bin/bash pid=$$ # get script's pid to var $pid echo $pid # add -A - show all option for cat cat -A /proc/$pid/cmdline 

there ya go, now we can see delimeters^@:

$ ./test 1 '2 3' 4 13002 /bin/bash^@./test^@1^@2 3^@4^@ 

lets add some more hacking to get the name:

#!/bin/bash pid=$$ # get script's pid to var $pid # tr to change '^@' into ' ' and awk to print 2nt column(script name) name=$(cat -A /proc/$pid/cmdline | tr '^@' ' ' | awk '{print $2}') echo " pid: $pid name: $name " 

result:

$ ./test 1 '2 3' 4 pid: 13091 name: ./test 

echo "You are running $0"