awk print full line before the matched pattern-CodePudding

so I am aware there are numerous posts trying to solve this question; however, I have had not succeeded in achieving that! So I have a script that tells me the directory from where my bash script is located which I took from another post:

#SCRIPT_DIR="$( cd -- "$( dirname -- "${BASH_SOURCE[0]}" )" &> /dev/null && pwd )"
/home/vMX-ENV/vMX-21.1R1/scripts/boot

I am trying to remove the /scripts/boot directory from the result so it would end up being:

/home/vMX-ENV/vMX-21.1R1/

I am trying to use awk as my solution, I know there are other commands like grep or perl but I rather use awk just to keep it consistent. Does anyone know how can I achieve that? I have had so many unsuccessful attempts

Someone suggested on another post to use

awk '/foo/{if (a && a !~ /foo/) print a; print} {a=$0}' file

But I couldn't make it work. Additionally, I have not much experience with bash so I couldn't figure out a way to solve it. Any help?

CodePudding user response：

Your question isn't clear but maybe this is what you're trying to do, i.e. just populate SCRIPT_DIR with the path to the directory 2 levels above the one where your script exists?

SCRIPT_DIR="$( cd -- "$( dirname -- "${BASH_SOURCE[0]}" )/../.." &> /dev/null && pwd )"

CodePudding user response：

Here is an AWK command which deletes the trailing part:

$ echo /home/vMX-ENV/vMX-21.1R1/scripts/boot | awk 'match($0, "/scripts/boot$") { print substr($0, 1, RSTART - 1) }'
/home/vMX-ENV/vMX-21.1R1

CodePudding user response：

Instead of using awk, you could use sed, as in:

echo /home/vMX-ENV/vMX-21.1R1/scripts/boot | sed 's/scripts\/boot$//'

This will give you as output:

/home/vMX-ENV/vMX-21.1R1/