How to make a script that will make another script in the same folder executable?

Question

How to make a script that will make another script in the same folder executable? I don't want to worry about path since it will be in the same directory. So it should be something like(i'm trying to make D3GO script executable):

#!/bin/bash    

chmod +x "$pwd"/D3GO

I'm very new to scripts so pls don't laugh :D Thanks!

Answer 1

When you run a bash script, all of it's commands run in the directory where the script is invoked from, including pwd. For example:

#!/bin/bash

pwd

If you run this from /whatever/path, it will display /whatever/path, no matter where the script is placed.

If you want to operate on files in the directory where you are running the script from, you can omit pwd like this:

chmod +x D3GO

This will be interpreted as chmod +x /whatever/path/D3GO if you run it from /whatever/path, no matter where the script is actually placed.

To operate in the directory where the script is located, you can use the $BASH_SOURCE variable. It is an array, and it's first element is the path to the script itself (i.e. /script/directory/myscript.sh). You can combine this with the dirname command to get just the directory:

chmod +x `dirname ${BASH_SOURCE[0]}`/D3GO

This will look for D3GO in the directory where this script is placed instead of where it's invoked from. (Thanks to @muru for pointing out my mistake.)

Answer 2

You were super close!

#!/bin/bash    

chmod +x D3GO