How to cut the first Sunday to Saturday of each month in a year?

I suggest the following alternative to awk:

#! /usr/bin/env bash
for month in {01..03}; do
  for day in {01..13}; do
    date -d "2020-$month-$day" '+%A %F'
  done |
  grep -A6 -m1 -F Sunday
done

The script is not very efficient, but does the job. For each month, we simply print the dates of the 13 first days in that month. We know that the green zone has to be in that area, therefore we do not need the remaining days of the month.
The date format is Weekday YYYY-MM-DD. We use grep to find and print the first Sunday, print the 6 days behind that Sunday (-A6) and exit because we limited the search to one match (-m1).
The procedure described above is done for each of the months 1 to 3.


With Perl, using DateTime

use warnings;
use strict;
use feature 'say';

use DateTime;

my $dt = DateTime->new(year => 2020, month => 1, day => 1); 

my $first_sunday = 7 - $dt->day_of_week + 1;  # day of month for first Sun

while (1) { 
    my $day = $dt->day; 
    if ($day >= $first_sunday and $day < $first_sunday + 7) { 
        say $dt->ymd, " (", $dt->day_abbr, ")";
    }
} 
continue { 
    $dt->add(days => 1); 
    if ($dt->day == 1) {  # new month
        last if $dt->month > 3;
        $first_sunday = 7 - $dt->day_of_week + 1;
    }   
}

This keeps a state (on the first in a month in finds out what day the first Sunday is), what is quite suitable if the program is meant to generate and go through all dates from the span of interest.

On the other hand, the program may need to check for a given day; perhaps it runs daily and needs to check for that day. Then it is simpler to see whether the day is between the first and second Sunday in the month

my $dt = DateTime->today;

while ( $dt->add(days => 1)->month <= 3) {

    if ($dt->day_of_week == 7) {           # it's a Sunday
        if ($dt->weekday_of_month == 1) {  # first Sunday in the month
            say $dt->ymd, " (", $dt->day_abbr, ")";
        }
    } 
    else {
        my $sdt = $dt->clone;                # preserve $dt
        $sdt->subtract( $dt->day_of_week );  # drop to previous Sunday
        if ($sdt->weekday_of_month == 1) {   # was first Sunday in the month
            say $dt->ymd, " (", $dt->day_abbr, ")";
        }
    }
}

The while loop around the code is there to facilitate a check.

For days other than Sunday we drop to the past Sunday, to check whether that was the first Sunday in the month. If so, then our day is within the required interval. If the day is a Sunday we only need to check whether it is the first one in the month.

The code can be made a bit more efficient and concise if that matters

if ( (my $dow = $dt->day_of_week) == 7) { 
    if ($dt->weekday_of_month == 1) {
        say $dt->ymd, " (", $dt->day_abbr, ")";
    }
}   
elsif ( $dt->clone->subtract(days => $dow)->weekday_of_month == 1 ) { 
    say $dt->ymd, " (", $dt->day_abbr, ")";
}

... on the account of readability.


Here's a simple way to get GNU awk to create a list of dates and day names for any given year:

$ cat tst.awk
BEGIN {
    year = (year == "" ? 2020 : year)
    beg = mktime(year " 1 1 12 0 0")
    for (i=0; i<=400; i++) {
        dateday = strftime("%F %A", beg+24*60*60*i)
        split(dateday,d,/[ -]/)
        if ( d[1] != year ) {
            break
        }
        print d[1], d[2], d[3], d[4]
    }
}

.

$ awk -f tst.awk | head -20
2020 01 01 Wednesday
2020 01 02 Thursday
2020 01 03 Friday
2020 01 04 Saturday
2020 01 05 Sunday
2020 01 06 Monday
2020 01 07 Tuesday
2020 01 08 Wednesday
2020 01 09 Thursday
2020 01 10 Friday
2020 01 11 Saturday
2020 01 12 Sunday
2020 01 13 Monday
2020 01 14 Tuesday
2020 01 15 Wednesday
2020 01 16 Thursday
2020 01 17 Friday
2020 01 18 Saturday
2020 01 19 Sunday
2020 01 20 Monday

I'm starting at noon and looping from 0 to 400 days and breaking when the year changes just so I don't have to try to accommodate DST or leap years or leap seconds in the determination of days in the year in a more accurate calculation.

Just add some code to test for the current month being different from the previous and the current day name being a Sunday and print 7 days starting there, e.g.:

$ cat tst.awk
BEGIN {
    year = (year == "" ? 2020 : year)
    beg = mktime(year " 1 1 12 0 0")
    for (i=0; i<=400; i++) {
        dateday = strftime("%F %A", beg+24*60*60*i)
        split(dateday,d,/[ -]/)
        if ( d[1] != year ) {
            break
        }
        dayName[d[2]+0][d[3]+0] = d[4]
    }
    for (monthNr=1; monthNr<=3; monthNr++) {
        for (dayNr=1; dayNr in dayName[monthNr]; dayNr++) {
            if (dayName[monthNr][dayNr] == "Sunday") {
                for (i=0; i<7; i++) {
                    printf "%s %04d-%02d-%02d\n", dayName[monthNr][dayNr+i], year, monthNr, dayNr+i
                }
                break
            }
        }
    }
}

.

$ awk -f tst.awk
Sunday 2020-01-05
Monday 2020-01-06
Tuesday 2020-01-07
Wednesday 2020-01-08
Thursday 2020-01-09
Friday 2020-01-10
Saturday 2020-01-11
Sunday 2020-02-02
Monday 2020-02-03
Tuesday 2020-02-04
Wednesday 2020-02-05
Thursday 2020-02-06
Friday 2020-02-07
Saturday 2020-02-08
Sunday 2020-03-01
Monday 2020-03-02
Tuesday 2020-03-03
Wednesday 2020-03-04
Thursday 2020-03-05
Friday 2020-03-06
Saturday 2020-03-07

There are slightly more efficient ways to do it but the above is clear and simple and will run in the blink of an eye.

Tags:

Bash

Perl

Awk

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