Store variable from output bash

Question

Got a bash script to create a backup of some files, it is set in Cron.

The bash command, including output, is:

[vagrant@localCentOS71 ~]$ sudo create_backup.sh
Check if all is good ...
done
Doing it ..
Backup file is: 1482419092_backup.tar ... Yeah!
done

How can I store 1482419092_backup.tar in a variable, for later use, in the same shell command.

The goal is to create a symlink to a folder of the just created backup file, in one shell command.

So far i've got:

create_backup.sh && ln -s $BACKUP_FILE /folder/here

Answer 1

It would be much simpler to just edit the script, but if that's not a possibility, you can do:

BACKUP_FILE=$(create_backup.sh | grep -oP 'Backup file is:\s*\K\S+') && 
ln -s $BACKUP_FILE /folder/here

Answer 2

Your shell script is executed by another shell which is a sub process.

There is no direct way to pass a variable from the child process to its parent.

One method could be to execute the script with your current shell with the dot (".") operator, but any call to exit will stop your current shell and you will inherit all the variables of your script :

. ./create_backup.sh && ln -s $BACKUP_FILE /folder/here

A better way is to use file to store your variable.

At some point in your script you save the value of the variable :

declare -p BACKUP_FILE > /tmp/somefile.sh

or :

echo BACKUP_FILE=\""$BACKUP_FILE"\" > /tmp/somefile.sh

Then you can execute :

if ./create_backup.sh
then . /tmp/somefile.sh
     ln -s $BACKUP_FILE /folder/here
fi

Another way without files, could be to read from the stdin of the script :

OUTSCRIPT=` ./create_backup.sh | grep 'BACKUP_FILE=' `

Here is an another solution : the parenthesis "()" creates a sub-shell to protect you from the exit calls and unwanted variables of the script AND you don't have to modify the script :

if ( . ./create_backup.sh && declare -p BACKUP_FILE > /tmp/somefile )
then . /tmp/somefile
     ln -s $BACKUP_FILE /folder/here
fi