pwd command with shell script

Question

I am calling a script as below

---

directory path : /user/local/script/print_path.sh

var_path=`pwd`
echo $var_path

The above script is calling as below directory path : /user/local/callPscript/call.sh

`/user/local/script/print_path.sh`

---

I want the out put as below :

/user/local/script/

But it gives the output :

/user/local/callPscript/

---

i.e. the pocation of the script is called. How can I make it to the scripts home directory path?

Answer 1

After some weeks of Bash programming, this has emerged as the standard solution:

directory=$(dirname -- $(readlink -fn -- "$0"))

$0 is the relative path to the script, readlink -f resolves that into an absolute path, and dirname strips the script filename from the end of the path.

A safer variant based on the completely safe find:

directoryx="$(dirname -- $(readlink -fn -- "$0"; echo x))"
directory="${directoryx%x}"

This should be safe with any filename - $() structures remove newlines at the end of the string, which is the reason for the x at the end.

Answer 2

May be this can help you.

var_path=$PWD  
echo $var_path