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Full Discussion: Shell Script for formatted output
Top Forums Shell Programming and Scripting Shell Script for formatted output Post 303001171 by rbatte1 on Friday 28th of July 2017 07:48:08 AM
Old 07-28-2017
A few pointers to tidy up first:-
  • Please wrap CODE tags [CODE] & [/CODE] around code, files, input & output/errors.
  • The append to feoj.txt can be added to the end of the loop
  • There is no need to use cat to pipe the file, just name the file as part of the command
  • Putting the awk before the sort will mean sort has less processing to do.
  • If you don't need to store the file, you can avoid the IO altogether.
  • What are the full filenames for the log files? The logfiles should be in order if they have a correct timestamp in the name.
    If they are just aged my moving them through the suffixes of -1, -2, -3 etc., then we can adjust for that.

Doing just these (but not removing the final sort) gives me this:-
Code:
#!bin/ksh

job_name=$1
echo $job_name

path="/Autosys/CA/UnicenterAutoSysJM/autouser.ACE/out"

cd $path                        # ..... or 'pushd $path' if you prefer so you can 'popd' later if needed

for filename in event_demon.ACE*
do
   zgrep $job_name $filename  | sed -e 's/\[/ /' -e 's/\]/ /'
done  | awk '/START|SUCCESS/ { print $1,$2}' | sort -k1


Now that it's a bit neater, where is it going wrong?


Kind regards,
Robin
Last edited by rbatte1; 07-28-2017 at 09:22 AM..
This User Gave Thanks to rbatte1 For This Post:
Sandeep Behera
 

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