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Full Discussion: How to find second and fourth Monday of the month?
Top Forums UNIX for Dummies Questions & Answers How to find second and fourth Monday of the month? Post 302948544 by Prathmesh on Tuesday 30th of June 2015 10:18:14 AM
Old 06-30-2015
Quote:
Originally Posted by rbatte1
Are you trying to determine if the script is started on the correct day, i.e. 2nd & 4th Monday only?

If so, start it every day and have a section to test the date at the top similar to this:-
Code:
date '+%d %w'|read dayofmonth dayofweek
((weekofmonth=1+$dayofmonth/7))

if [ $dayofweek -ne 1 -o \( $weekofmonth -ne 2 -a $weekofmonth -ne 4 \) ]
then
   printf "Not running today.\n"
   exit 0
fi

# Now able to run main script

Apologies if I have missed the point, but I hope that this helps.

Robin
Thanks Robin. That's what the intention is.

Could you please advise why you are adding 1 in the below code:
Code:
((weekofmonth=1+$dayofmonth/7))

Is it related to day of the week? Or It will be same irrespective of whether the month starts on Mon,Tue,Wed,Thu,Fri,Sat or Sun?

---------- Post updated at 07:48 PM ---------- Previous update was at 07:45 PM ----------

Quote:
Originally Posted by RudiC
Looks like your shell doesn't zero pad the MONTH sequence as does my bash. You may want to consider the various options that your shell offers for "brace expansion". Otherwise, prepend single digit MONTHs with a "0" by shell means.

---------- Post updated at 09:40 ---------- Previous update was at 09:39 ----------

or, use seq -w 1 12.
Thanks RudiC.
It is working now.

Code:
$ for MONTH in `seq -w 1 12`
>     do printf "%2s.%2s.%4s      %2s.%2s.%4s\n" \
>                 $(($(date -d"2015${MONTH}01" +"16 - %u - (%u==1?7:0)"))) $MONTH 2015 \
>                 $(($(date -d"2015${MONTH}01" +"30 - %u - (%u==1?7:0)"))) $MONTH 2015
> done
12.01.2015      26.01.2015
 9.02.2015      23.02.2015
 9.03.2015      23.03.2015
13.04.2015      27.04.2015
11.05.2015      25.05.2015
 8.06.2015      22.06.2015
13.07.2015      27.07.2015
10.08.2015      24.08.2015
14.09.2015      28.09.2015
12.10.2015      26.10.2015
 9.11.2015      23.11.2015
14.12.2015      28.12.2015
$

I have also managed to do the same using below code.
Code:
$ for MONTH in `seq -w 1 12`
> do
> cal $MONTH $YEAR | tail -n +3 | cut -c4-5 | tr -s '\n' ','  | sed 's/[ \t]*//g' | awk -F "," -v M=$MONTH -v Y=$YEAR '{if($1>7 || $1==""){printf "%2s.%2s.%4s\t%2s.%2s.%4s\n",$3,M,Y,$5,M,Y} else {printf "%2s.%2s.%4s\t%2s.%2s.%4s\n",$2,M,Y,$4,M,Y}}'
> done
12.01.2015      26.01.2015
 9.02.2015      23.02.2015
 9.03.2015      23.03.2015
13.04.2015      27.04.2015
11.05.2015      25.05.2015
 8.06.2015      22.06.2015
13.07.2015      27.07.2015
10.08.2015      24.08.2015
14.09.2015      28.09.2015
12.10.2015      26.10.2015
 9.11.2015      23.11.2015
14.12.2015      28.12.2015
$

Let me know if this code is fine.
Last edited by Prathmesh; 06-30-2015 at 11:21 AM.. Reason: Added my code
 

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