How to Access the Base Filename of a File You Are Sourcing in Bash

How can you access the base filename of a file you are sourcing in Bash?

See Getting the source directory of a Bash script from within, particularly the comment regarding the BASH_SOURCE variable.

Summary: SCRIPT_NAME=$(basename ${BASH_SOURCE[0]})

How to find out name of script called (sourced) by another script in bash?

Change file2.sh to:

#!/bin/bash
echo "from file2: ${BASH_SOURCE[0]}"

Note that BASH_SOURCE is an array variable. See the Bash man pages for more information.

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).

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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.

How do I get the directory where a Bash script is located from within the script itself?

#!/usr/bin/env bash

SCRIPT_DIR=$( cd -- "$( dirname -- "${BASH_SOURCE[0]}" )" &> /dev/null && pwd )

is a useful one-liner which will give you the full directory name of the script no matter where it is being called from.

It will work as long as the last component of the path used to find the script is not a symlink (directory links are OK). If you also want to resolve any links to the script itself, you need a multi-line solution:

#!/usr/bin/env bash

SOURCE=${BASH_SOURCE[0]}
while [ -L "$SOURCE" ]; do # resolve $SOURCE until the file is no longer a symlink
  DIR=$( cd -P "$( dirname "$SOURCE" )" >/dev/null 2>&1 && pwd )
  SOURCE=$(readlink "$SOURCE")
  [[ $SOURCE != /* ]] && SOURCE=$DIR/$SOURCE # if $SOURCE was a relative symlink, we need to resolve it relative to the path where the symlink file was located
done
DIR=$( cd -P "$( dirname "$SOURCE" )" >/dev/null 2>&1 && pwd )

This last one will work with any combination of aliases, source, bash -c, symlinks, etc.

Beware: if you cd to a different directory before running this snippet, the result may be incorrect!

Also, watch out for $CDPATH gotchas, and stderr output side effects if the user has smartly overridden cd to redirect output to stderr instead (including escape sequences, such as when calling update_terminal_cwd >&2 on Mac). Adding >/dev/null 2>&1 at the end of your cd command will take care of both possibilities.

To understand how it works, try running this more verbose form:

#!/usr/bin/env bash

SOURCE=${BASH_SOURCE[0]}
while [ -L "$SOURCE" ]; do # resolve $SOURCE until the file is no longer a symlink
  TARGET=$(readlink "$SOURCE")
  if [[ $TARGET == /* ]]; then
    echo "SOURCE '$SOURCE' is an absolute symlink to '$TARGET'"
    SOURCE=$TARGET
  else
    DIR=$( dirname "$SOURCE" )
    echo "SOURCE '$SOURCE' is a relative symlink to '$TARGET' (relative to '$DIR')"
    SOURCE=$DIR/$TARGET # if $SOURCE was a relative symlink, we need to resolve it relative to the path where the symlink file was located
  fi
done
echo "SOURCE is '$SOURCE'"
RDIR=$( dirname "$SOURCE" )
DIR=$( cd -P "$( dirname "$SOURCE" )" >/dev/null 2>&1 && pwd )
if [ "$DIR" != "$RDIR" ]; then
  echo "DIR '$RDIR' resolves to '$DIR'"
fi
echo "DIR is '$DIR'"

And it will print something like:

SOURCE './scriptdir.sh' is a relative symlink to 'sym2/scriptdir.sh' (relative to '.')
SOURCE is './sym2/scriptdir.sh'
DIR './sym2' resolves to '/home/ubuntu/dotfiles/fo fo/real/real1/real2'
DIR is '/home/ubuntu/dotfiles/fo fo/real/real1/real2'

Unix shell script find out which directory the script file resides?

In Bash, you should get what you need like this:

#!/usr/bin/env bash

BASEDIR=$(dirname "$0")
echo "$BASEDIR"

Reliable way for a Bash script to get the full path to itself

Here's what I've come up with (edit: plus some tweaks provided by sfstewman, levigroker, Kyle Strand, and Rob Kennedy), that seems to mostly fit my "better" criteria:

SCRIPTPATH="$( cd -- "$(dirname "$0")" >/dev/null 2>&1 ; pwd -P )"

That SCRIPTPATH line seems particularly roundabout, but we need it rather than SCRIPTPATH=`pwd` in order to properly handle spaces and symlinks.

The inclusion of output redirection (>/dev/null 2>&1) handles the rare(?) case where cd might produce output that would interfere with the surrounding $( ... ) capture. (Such as cd being overridden to also ls a directory after switching to it.)

Note also that esoteric situations, such as executing a script that isn't coming from a file in an accessible file system at all (which is perfectly possible), is not catered to there (or in any of the other answers I've seen).

The -- after cd and before "$0" are in case the directory starts with a -.

How do I get `__FILE__` when sourcing a csh script

If you access $_ on the first line of the file, it will contain the name of the file if it's sourced. If it's run directly then $0 will contain the name.

#!/bin/tcsh
set called=($_)
if ($called[2] != "") echo "Sourced: $called[2]"
if ($0 != "tcsh") echo "Called: $0"

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