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How to find second and fourth Monday of the month?

 
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# 1  
Old 06-29-2015
How to find second and fourth Monday of the month?
Hi,

I have came across the scenario where, we have to run the script on second and fourth Monday of each month.

I have tried to search man page of date and also forum for it but, could not get any answer to this.

Can you please advise how can we get second and fourth Monday of the month?

Also, While searching for above problem, I have came across the thread Can we get every tuesday or monday's date for the current week which has following code to find Monday of the current week.

Code:
cal | awk -v d="$(date +"%m %d %y")" 'BEGIN{split(d,a)} $0 ~ int(a[2]) {print a[1] $2 a[3]}'

I am unable to understand that code and that thread is closed. Can someone please explain this code?

Thanks in advance for your help.
Last edited by rbatte1; 06-30-2015 at 07:11 AM.. Reason: Corrected link
# 2  
Old 06-29-2015
cal without any further parameters prints the current month in 4 - 5 lines plus two header lines. In these lines, the awk searches for today and, if found, prints $2, the second field, which is Monday as the weeks in cal start with Sunday.
# 3  
Old 06-29-2015
You probably want that script to be executed by cron. https://www.unix.com/man-page/opensolaris/1m/cron/

But also see the crontab entry: https://www.unix.com/man-page/centos/5/crontab/

Code:
*     *     *   *    *        command to be executed
-     -     -   -    -
|     |     |   |    |
|     |     |   |    +----- day of week (0 - 6) (Sunday=0)
|     |     |   +------- month (1 - 12)
|     |     +--------- day of        month (1 - 31)
|     +----------- hour (0 - 23)
+------------- min (0 - 59)

So this might be what you want: (untested)
Code:
* * * * 1/2,4 /path/to/script

This suggestion is based on what i read myself just now about cron.
hth
# 4  
Old 06-29-2015
With GNU date and a recent bash, try this:
Code:
for MONTH in {01..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

Feel free to adapt to other years.
# 5  
Old 06-29-2015
Quote:
Originally Posted by sea
You probably want that script to be executed by cron. https://www.unix.com/man-page/opensolaris/1m/cron/

But also see the crontab entry: https://www.unix.com/man-page/centos/5/crontab/

Code:
*     *     *   *    *        command to be executed
-     -     -   -    -
|     |     |   |    |
|     |     |   |    +----- day of week (0 - 6) (Sunday=0)
|     |     |   +------- month (1 - 12)
|     |     +--------- day of        month (1 - 31)
|     +----------- hour (0 - 23)
+------------- min (0 - 59)

So this might be what you want: (untested)
Code:
* * * * 1/2,4 /path/to/script

This suggestion is based on what i read myself just now about cron.
hth
Thanks for your reply. But, I do not want to use cron. I want to control second and fourth Monday through shell script.

---------- Post updated at 12:50 AM ---------- Previous update was at 12:48 AM ----------

Quote:
Originally Posted by RudiC
With GNU date and a recent bash, try this:
Code:
for MONTH in {01..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

Feel free to adapt to other years.
Thanks. I will try this. But, can you please explain the logic used here.
# 6  
Old 06-29-2015
date's %u format specifier supplies the day-of-week of the first day in the target month, which then is subtracted from 16 (for second week) or 30 (4. wk, experimental values). Should the First be a Monday, subtract another 7 days.
# 7  
Old 06-30-2015
Getting below errors, any suggestions?
Code:
$ for MONTH in {01..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
date: invalid date `2015101'
date: invalid date `2015101'
 0. 1.2015       0. 1.2015
date: invalid date `2015201'
date: invalid date `2015201'
 0. 2.2015       0. 2.2015
date: invalid date `2015301'
date: invalid date `2015301'
 0. 3.2015       0. 3.2015
date: invalid date `2015401'
date: invalid date `2015401'
 0. 4.2015       0. 4.2015
date: invalid date `2015501'
date: invalid date `2015501'
 0. 5.2015       0. 5.2015
date: invalid date `2015601'
date: invalid date `2015601'
 0. 6.2015       0. 6.2015
date: invalid date `2015701'
date: invalid date `2015701'
 0. 7.2015       0. 7.2015
date: invalid date `2015801'
date: invalid date `2015801'
 0. 8.2015       0. 8.2015
date: invalid date `2015901'
date: invalid date `2015901'
 0. 9.2015       0. 9.2015
12.10.2015      26.10.2015
 9.11.2015      23.11.2015
14.12.2015      28.12.2015
$

 
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