passing $var(korn shell var) to awk

how do you pass $var to awk:- see below

for var in `ls *.gz`
echo $var
zcat $var | grep "HTML" | awk '{printf
("%s===>%s,(need $var here),$0)}'

how do i get $var into awk as it contains the file name, the output should look like:

filename.gz===>HTML record no 50

any takers??
Who is Participating?
I wear a lot of hats...

"The solutions and answers provided on Experts Exchange have been extremely helpful to me over the last few years. I wear a lot of hats - Developer, Database Administrator, Help Desk, etc., so I know a lot of things but not a lot about one thing. Experts Exchange gives me answers from people who do know a lot about one thing, in a easy to use platform." -Todd S.

I don't use ksh but it should be no different than bash for thsi application.

for var in a b c d e
  echo testing | awk '{printf"%s: %s\n",$0,a}' a=$var

testing: a
testing: b
testing: c
testing: d
testing: e

You pass the variables in as arguments to awk after the instructions. Note no "$" for the variable naming in the awk instructions.



Experts Exchange Solution brought to you by

Your issues matter to us.

Facing a tech roadblock? Get the help and guidance you need from experienced professionals who care. Ask your question anytime, anywhere, with no hassle.

Start your 7-day free trial
e2venuAuthor Commented:
that is what i needed, great, simple ones that sting the most
It's more than this solution.Get answers and train to solve all your tech problems - anytime, anywhere.Try it for free Edge Out The Competitionfor your dream job with proven skills and certifications.Get started today Stand Outas the employee with proven skills.Start learning today for free Move Your Career Forwardwith certification training in the latest technologies.Start your trial today
Web Languages and Standards

From novice to tech pro — start learning today.