Remove all but the latest X files from sftp via bash-script - linux

Question

I have a working bash script to create backups and upload them as a tar archive to a remote sftp server.
After the upload, the script should remove all but the latest 20 backup files. I can't use any, pipe, grep, whatever on the sftp. Also I don't get the file-listing result handled in my bash-script.
export SSHPASS=$(cat /etc/backup/pw)
SFTPCONNECTION=$(cat /etc/backup/sftp-connection)
sshpass -e sftp $SFTPCONNECTION - << SOMEDELIMITER
ls -lt backup-*.tar
quit
SOMEDELIMITER
There is this nice oneliner, but I did not figure out how the use it in my case (sftp).

Answer 1

This script deletes all tar files in the given directory except the last 20 ones. The -t flag sorts by time & date. The <<< redirect expands $RESULT feed's it into the stdin of the while loop. I'm not entirely pleased with it as it has to create multiple connections, but with sftp I don't believe there is another way.
RESULT=`echo "ls -t path/to/old_backups/" | sftp -i ~/.ssh/your_ssh_key user#server.com | grep tar`
i=0
max=20
while read -r line; do
(( i++ ))
if (( i > max )); then
echo "DELETE $i...$line"
echo "rm $line" | sftp -i ~/.ssh/your_ssh_key user#server.com
fi
done <<< "$RESULT"

Answer 2

Thanks to codelitt I went with this solution:
export SSHPASS=$(cat /etc/backup/pw)
SFTPCONNECTION="username#host"
RESULT=`echo "ls -tl backup*.tar" | sshpass -e sftp $SFTPCONNECTION | grep -oP "backup.*\.tar" `
i=0
max=24
while read -r line; do
# echo "$line "
(( i++ ))
if (( i > max )); then
echo "DELETE $i...$line"
echo "rm $line" | sshpass -e sftp $SFTPCONNECTION
fi
done <<< "$RESULT"
It's a slight modification of his version:
it counts/removes only files named backup*.tar
it uses ls -l (for line based listings)
I had to use sshpass instead of a certificate-based authentication. The sftp password is inside /etc/backup/pw