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The accepted answer to Getting the source directory of a Bash script from within addresses getting the path of the script via dirname $0 , which is fine, but that may return a relative path (like . ), which is a problem if you want to change directories in the script and have the path still point to the script's ...
To determine the exact location of your current directory within the file system, go to a shell prompt and type the command pwd. This tells you that you are in the user sam's directory, which is in the /home directory. The command pwd stands for print working directory.
For Bash, you simply need to add the line from above, export PATH=$PATH:/place/with/the/file, to the appropriate file that will be read when your shell launches. There are a few different places where you could conceivably set the variable name: potentially in a file called ~/. bash_profile, ~/. bashrc, or ~/.
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`
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