Determine the path of the executing BASH script [duplicate]

Question

For the relative path (i.e. the direct equivalent of Windows' %~dp0):

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

For the absolute, normalized path:

MY_PATH=$(dirname "$0")            # relative
MY_PATH=$(cd "$MY_PATH" && pwd)    # absolutized and normalized
if [[ -z "$MY_PATH" ]] ; then
  # error; for some reason, the path is not accessible
  # to the script (e.g. permissions re-evaled after suid)
  exit 1  # fail
fi
echo "$MY_PATH"

Assuming you type in the full path to the bash script, use $0 and dirname, e.g.:

#!/bin/bash
echo "$0"
dirname "$0"

Example output:

$ /a/b/c/myScript.bash
/a/b/c/myScript.bash
/a/b/c

If necessary, append the results of the $PWD variable to a relative path.

EDIT: Added quotation marks to handle space characters.

Contributed by Stephane CHAZELAS on c.u.s. Assuming POSIX shell:

prg=$0
if [ ! -e "$prg" ]; then
  case $prg in
    (*/*) exit 1;;
    (*) prg=$(command -v -- "$prg") || exit;;
  esac
fi
dir=$(
  cd -P -- "$(dirname -- "$prg")" && pwd -P
) || exit
prg=$dir/$(basename -- "$prg") || exit 

printf '%s\n' "$prg"

Vlad's code is overquoted. Should be:

MY_PATH=`dirname "$0"`
MY_PATH=`( cd "$MY_PATH" && pwd )`

echo Running from `dirname $0`