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

bash, directory

Solution

#!/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'

Problem

How do I get the path of the directory in which a Bash script is located, inside that script? I want to use a Bash script as a launcher for another application. I want to change the working directory to the one where the Bash script is located, so I can operate on the files in that directory, like so: ``` $ ./application ```

Original source

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