Counting records with AWK

Shell Programming and Scripting
1

I've been working with an awk script and I'm wondeing id it's possible to count records in a file which DO NOT contain, in this instance fields 12 and 13.

With the one script I am wanting to display the count for the records WITH fields 12 and 13 and a seperate count of records WITHOUT fields 12 and 13.

Here's my script:

BEGIN {
print
print "User Process ID Time Active"
print
}
$4 $6 $7 $8 $9 $10 $11 {
cmbaselicences ++
}
$12 $13 {
changebaselicences ++
}
{
print $4, $6, $7, $8, $9, $10, $11, $12, $13
}
END {
printf "\n\nUsers of SYNERGY-CMBase: (Total of 12 licenses issued; Total of %-2d", cmbaselicences
print " licences in use)"
printf "\n\nUsers of SYNERGY-ChangeBase: (Total of 11 licenses issued; Total of %-2d", changebaselicences
print " licences in use)\n"
print
}

and the file with the records..:

ccm_root phys-agsdev /dev/tty ChangeAdmin (v1.0) (Phys-agsdev/19353 201), start Tue 1/20 6:30 (linger: 1800)
ccm_root phys-agsdev /dev/pts/58 u414020 (v1.0) (Phys-agsdev/19353 745), start Tue 1/20 18:43

Fields 12 and 13 contain the "(linger: 1800)" entry.

Cheers

2

Replace this:

$4 $6 $7 $8 $9 $10 $11 {
cmbaselicences ++
}
$12 $13 {
changebaselicences ++
}

with:

NF==11{cmbaselicences++}
NF==13{changebaselicences++}

Regards

3 
awk'
NF > 11 { ++changebaselicences }
NF < 12{ ++cmbaselicences }
END { ..... }
'