Uploading current day's backup instead of all files via Cron jobs - bash

Question

I have a VPS running on Centos 7, and created a cron job to dump my database (Sql 8.0) and to create a tar to backup my entire site's files and this goes on everyday
I want to create another bash / cron job to connect to my backup server and upload those backup files stored on my VPS.
The problem is, I can't get it to upload only the newest files, not the entire files as there will be 7 backups every week.
I want it to only upload today's files, not all available files.
Should I use rsync ?
Here is my bash so far:
#!/bin/sh
USERNAME="ftp user"
PASSWORD="ftp password"
SERVER="IP or domain"
# local directory to pickup *.tar.gz file
FILE="/path/"
# remote server directory to upload backup
BACKUPDIR="/pro/backup/sql"
# login to remote server
ftp -n -i $SERVER <<EOF
user $USERNAME $PASSWORD
cd $BACKUPDIR
mput $FILE/*.tar.gz
quit
EOF

Answer 1

You can use find with -ctime to search for .tar.gz files changed in the last 7 days and then loop on the results, ftping each. Using this logic with your existing solution:
#!/bin/sh
USERNAME="ftp user"
PASSWORD="ftp password"
SERVER="IP or domain"
# local directory to pickup *.tar.gz file
FILE="/path/"
# remote server directory to upload backup
BACKUPDIR="/pro/backup/sql"
while read fil;
do
# login to remote server
ftp -n -i $SERVER <<EOF
user $USERNAME $PASSWORD
cd $BACKUPDIR
mput "$FILE/$fil"
quit
EOF
done < "$(find $FILE -ctime -7 -name "*.tar.gz")"