passing $var(korn shell var) to awk

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

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

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??
e2venuAsked:
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s_franklinCommented:
I don't use ksh but it should be no different than bash for thsi application.

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

Output:
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.

Steve


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e2venuAuthor Commented:
that is what i needed, great, simple ones that sting the most
0
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