[BASH] Gawk + MYSQL script

Shell Programming and Scripting
1

Hello!

I've got script to write. It should read databases (names, volumes) from table testdatabase and compares it to actually existing databases in /var/lib/mysql/. If there is no informations about database in table - we should see information "There is no declared informations about database in table testdatabase." And if volume of database is too big - we should see information "Database database has too big actual size volume, it exceeds declared size in table testdatabase."

My problem is that even if there is existing database in /var/lib/mysql/ and I put informations about it in table testdatabase - I see statement - "There is no declared...".

Do you have any idea what am I doing wrong?

cd /var/lib/mysql/ || exit 1

gawk '
BEGIN {
   cmd = "mysql -u root -ppassword knowdatabases \"SELECT name, volume, date, description FROM testdatabase\""
   while ( cmd | getline ) {
      DATABASE[$1,1] = $2;
      DATABASE[$1,2] = $1;
   }
   close(cmd)
   
   cmd = "find /var/lib/mysql/ -maxdepth 1 -mindepth 1 -type d | xargs du -sm"
   while ( cmd | getline ) {
      VOL = $1;
      NAM = $2;
      if ( ! DATABASE[NAM,1] ) {
         print "There is no declared informations about " NAM " in table testdatabase."
      continue
      }
      if ( VOL > DATABASE[NAM,1] ) {
         print "Database " NAM " has too big actual size (" VOL "), it exceeds declared size in table testdatabase."
      continue
      }
   }
   close(cmd)
}
' /dev/null
2

You could do much more is SQL, with case. Part of the task is to compare lists, which should say if either list is short, something comm is good at. You could sort and rely on order to merge the lists.