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Shell Programming and Scripting

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Top Forums Shell Programming and Scripting awk experts please help
# 8  
Old 01-16-2013
My bad, you have to modify both places:-
Code:
nawk '/PIN_FLD_UNIQUE_ID/ {
 uid=$NF;gsub(/"/,"",uid);
 print uid,dt,dt,per;
} /PIN_FLD_PERCENT/ {
 per=$NF;
} /UnknownProgramName/ {
 dt=sprintf("%s %s",$2,$3);
 gsub(/\..*/,"",dt);
}' OFS=, filename | nawk '!a[$0]++' > output_filename

I hope it helps.
This User Gave Thanks to Yoda For This Post:
mirwasim
# 9  
Old 01-16-2013
Yes it worked.
Last edited by mirwasim; 01-17-2013 at 03:38 AM..
# 10  
Old 01-16-2013
I'm not sure what you mean by 80%, Since you have the formatted output in a file, I believe this will be something that you can work & figure out yourself.
# 11  
Old 01-16-2013
Code:
1234, 2012-12-02 23:16:17 , 2012-12-02 23:16:17

Last edited by mirwasim; 01-17-2013 at 03:38 AM..
# 12  
Old 01-16-2013
It is pretty much straightforward, before printing the values you have to put a condition:
Code:
if ((per==80)||(per==100)||(per>100))
 print uid,dt,dt,per;

Change the condition as per your requirement.
# 13  
Old 01-16-2013
Thanks ...very much helpful

---------- Post updated at 03:06 AM ---------- Previous update was at 03:02 AM ----------


Code:
nawk '/PIN_FLD_UNIQUE_ID/ { 

 uid=$NF;gsub(/"/,"",uid);
 print uid,dt,dt,per;
} /PIN_FLD_PERCENT/ {
 per=$NF;
} /UnknownProgramName/ {
 dt=sprintf("%s %s",$2,$3);
 gsub(/\..*/,"",dt);

}' OFS=, $1 | nawk '!a[$0]++'
if ((per==80)||(per==100)||(per>100))
 print uid,dt,dt,per; > output_filename

Last edited by mirwasim; 01-17-2013 at 03:42 AM..
# 14  
Old 01-16-2013
Here:
Code:
nawk '/PIN_FLD_UNIQUE_ID/ {
 uid=$NF;gsub(/"/,"",uid);
 if ((per==80)||(per==100)||(per>100))
   print uid,dt,dt,per;
} /PIN_FLD_PERCENT/ {
 per=$NF;
} /UnknownProgramName/ {
 dt=sprintf("%s %s",$2,$3);
 gsub(/\..*/,"",dt);
}' OFS=, filename | nawk '!a[$0]++'

This User Gave Thanks to Yoda For This Post:
mirwasim
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