awk: find a pattern then return the previous line

Here's an awk script (it's a bit long to call it a command) that looks for a pattern and then returns the previous line. I attempted this when someone told me it could'nt be done with awk - I just did it to prove a point and spent way too much time figuring it out!

Command line:

awk -v N=PATTERN -f getprevline1 testinput.dat

Contents of getprevline1:

BEGIN {
(getline line1 <= ARGV[1])
}
{
if ((getline line < ARGV[1]) && $0 ~ /N/ ) {
print line
}
}

You can see that we're using the -v flag again to define a variable called N with the value 'PATTERN'. We're also using the -f (lowercase) flag for the first time. This flag calls the awk command file getprevline1.

We're also bombarded with the getline awk command which is something that should'nt be dabbled with lightly. In one of the awk books I have it essentially says don't use getline until you've mastered everything else!