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For Args and Nawk

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# 1  
Old 12-19-2011
For Args and Nawk
I am trying to write a simple shell script that will take certain arguments (numerical values) and plug each one into a nawk command.

I thought I would need to use for args x y z but i get syntax errors:
Code:
for args 16 1 3 25 31 41
do
nawk -F, '{if($10==$ && $12~/CA[1-5]$/){SUM+=$13}}END{print(SUM)}' /opt/PlexDC/statsdir/PegCount/secondary/*20111217*
done

So the script is supposed to take each argument (16, 1, 3, 25, etc) and plug into the nawk command where it says "$10==$", and $ is supposed to be the argument.

Smilie

Heather
Last edited by radoulov; 12-19-2011 at 04:15 PM.. Reason: Code tags!
# 2  
Old 12-19-2011
You're close. Your for syntax is slightly off, and getting it "into" awk isn't always obvious.

Code:
for arg in 16 1 3 25 31 41
 do
 nawk -F, -v arg=$arg '{if($10==arg && $12~/CA[1-5]$/){SUM+=$13}}END{print(SUM)}' /opt/PlexDC/statsdir/PegCount/secondary/*20111217*
 done

EDIT: Forgot to mention that the -v arg=$arg assigns the shell variable's value to the awk variable allowing it to be referenced in the awk programme. I also changed 'args' to 'arg' as it holds only one value at a time and the plural (IMHO) is misleading.
Last edited by agama; 12-19-2011 at 01:46 PM.. Reason: clarification
# 3  
Old 12-19-2011
You haven't even fed 'args' into nawk anywhere. Usually you'd do so by nawk -v VARNAME="$shellvar"

so
Code:
for arg in 16 1 3 25 31 41
do
        nawk -F, -v ARG="$arg" '{if($10==ARG && $12~/CA[1-5]$/){SUM+=$13}}END{print(SUM)}' /opt/PlexDC/statsdir/PegCount/secondary/*20111217*
done

This will run awk 6 separate times, with a different value of ARG every time, and print all of their output directly into the terminal. Is this precisely what you want?
# 4  
Old 12-19-2011
Thank you!! Happy Holidays!

---------- Post updated at 01:25 PM ---------- Previous update was at 12:14 PM ----------

Ok, so another question...this command works getting yesterday's date calculated:

Code:
date | awk '{printf"%4d%2d%2d\n",$6,$2,($3-1)}' | sed 's/ /0/g'

But I want to get that calculation into the original command I had where the "*2011217*" is located (so I want my script to automatically run these summaries on yesterday's file, and not have to change the script each day for the new date).
Moderator's Comments:
Mod Comment
Please use code tags when posting data and code samples!
Last edited by vgersh99; 12-19-2011 at 04:15 PM.. Reason: code tags, please!
# 5  
Old 12-19-2011
Quote:
Originally Posted by he204035
Thank you!! Happy Holidays!

---------- Post updated at 01:25 PM ---------- Previous update was at 12:14 PM ----------

Ok, so another question...this command works getting yesterday's date calculated:

date | awk '{printf"%4d%2d%2d\n",$6,$2,($3-1)}' | sed 's/ /0/g'


But I want to get that calculation into the original command I had where the "*2011217*" is located (so I want my script to automatically run these summaries on yesterday's file, and not have to change the script each day for the new date).
No, it doesn't. What happens on the first of March?
Search the FAQs on how to do the date math.
# 6  
Old 12-19-2011
I figured it out!

This works!

Code:
date +%Y%m%d | awk '{printf"%8d\n",($1-1)}'

Thanks for your help.
Last edited by zxmaus; 12-19-2011 at 09:41 PM..
# 7  
Old 12-19-2011
I'll reiterate what vgersh99 asked before: what happens when the date command returns something like 20120101? 20120100 is an invalid date, so beware that your code doesn't handle edge conditions. In short you must do the math individually for year, month and day taking into account the number of days in each month, and leap year.
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