i use this script to get the time and date of back and fourth transactions for a particular execution id. I use a substr command on the 5th column to to cut the milli seconds off the time value. - otherwise the times would look like 08:30:04.235

grep <executionID> <auditfile> | awk '{ print $1, $2, $3, $4, substr($5,1,8}

FIX -> Mon 3/1/2010 08:30:04
FIX <- Mon 3/1/2010 08:32:36
FIX <- Mon 3/1/2010 08:35:08
FIX -> Mon 3/1/2010 08:35:27
FIX <- Mon 3/1/2010 08:37:02
FIX <- Mon 3/1/2010 08:39:37
FIX -> Mon 3/1/2010 08:39:42

anyhow - i append two sed commands to further clarify the direction of the message.

awk '{ print $1, $2, $3, $4, substr($5,1,8} | sed -e 's/->/ ->IN/g' | sed -e 's/<-/<-OUT/g'

FIX -> IN Mon 3/1/2010 08:30:04
FIX <- OUT Mon 3/1/2010 08:32:36
FIX <- OUT Mon 3/1/2010 08:35:08
FIX -> IN Mon 3/1/2010 08:35:27
FIX <- OUT Mon 3/1/2010 08:37:02
FIX <- OUT Mon 3/1/2010 08:39:37
FIX -> IN Mon 3/1/2010 08:39:42

I tried using an awk gsub () command within the string instead of the two seds, but it did not work:
awk '{ print gsub(/<regex>/, <replace with>,$1), $2, $3, $4, substr($5,1,8}

the sed works ok, but it would be cooler to make the replacement within the awk command:

gsub(/->/,-> IN, $1)

Is there a way where i could replace the value of the $1 column in the awk print string?

Your problem is that you're trying to use gsub in the print
statement. You can do two things:
a) use gensub instead of gsub
b) separate gsub from print.
In this case I'd prefer b) - it's cleaner & shorter.


Cheers,
Tink