Change The Call from a Bash Script in a Month to a Week

change the number of week by the range of dates in bash

This will work on linux with bash, assuming that by "week number", you mean the ISO week number. ISO weeks start on Monday and a week counts as part of a year if four of its days are in the year.

dates_for_week() {
  printf "January 1 $1 %+d days\n" {-3..368} |
    date +%G%V_%Y-%m-%d -f - |
    grep $(printf %d%02d $1 $2) |
    cut -c8-
}

EDIT: Here's a version which does less work to get the same result:

dates_for_week () { 
  printf "$1-01-01 +$2 weeks %+d days\n" {-10..2} |
    date +%V_%Y-%m-%d -f - |
    grep '^0\?'$2_ |
    cut -c4-
}

eg:

$ dates_for_week 2013 1
2012-12-31
2013-01-01
2013-01-02
2013-01-03
2013-01-04
2013-01-05
2013-01-06
$ dates_for_week 2013 46
2013-11-11
2013-11-12
2013-11-13
2013-11-14
2013-11-15
2013-11-16
2013-11-17
$ dates_for_week 2013 52
2013-12-23
2013-12-24
2013-12-25
2013-12-26
2013-12-27
2013-12-28
2013-12-29

How to send out current week in shell scripts

The statement

WEEK_NUMBER = %V

(with spaces) will try to run the WEEK_NUMBER program, passing = and %V as arguments.

What you need is what you have on the OUTPUT_SUMMARY line above, an assignment with no spaces:

WEEK_NUMBER=%V

In any case, that would simply set WEEK_NUMBER to the literal %V. Since those %-arguments are meant for the date command, you should probably use:

WEEK_NUMBER=$(date +%V)          # Store output of date week-no.
YEAR=$(date +%Y)                 # Store output of date year.
WORK_WEEK=${YEAR}.${WEEK_NUMBER} # Construct from variables.

Of course, if the only thing you want to use is the WORK_WEEK variable, it can be simplified quite a lot, as date allows you to insert arbitrary text into its output:

WORK_WEEK=$(date +%Y.%V)

This will also get rid of the potential bug when running your script very late on December 31st. In that case, it's possible for the week number is extracted in year N and the year extracted in year N+1. Unlikely, yes, but I'm paid to worry about such things :-)

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.

How to pass a range of dates through the for loop in bash?

Don't do it yourself. Dates are tricky, months have variable number of days, etc. Your script also needs additional handling of years. Let date handle calculations with dates.

for i in {1..7}; do ./scriptname.sh $(date --date="-$i days" "+%Y %m %d"); done

.g the week running from 29th March through 4th April

The way I would do it is to take 29th Match as number of seconds since epoch. Then add half a day so that leap seconds if any are not an isssue. Then just increment it by number of seconds in a day and use date to convert it back to a date.

then=$(date --date='2020/03/29 12:00:00' +%s);
for i in {0..6}; do 
     ./scriptname.sh $(date --date="@$((then + $i * 60 * 60 * 24))" "+%Y %m %d");
done

Alternatively, you can do it without then variable, but you have to remember that the number +0 immediately coming after date is interpreted as timezone.

for i in {0..6}; do ./scriptname.sh $(date --date="2020/03/29 12:00:00 +0 +$i days" '+%Y %m %d'); done

Because I like streaming-like parsing in bash, I would parse the output with xargs:

seq 6 | xargs -I{} date -d '2020/03/29 12:00:00Z +{} days' '+%Y %m %d' | xargs -n3 ./scriptname.sh

Get the date (a day before current time) in Bash

Sorry not mentioning I on Solaris system.
As such, the -date switch is not available on Solaris bash.

I find out I can get the previous date with little trick on timezone.

DATE=`TZ=MYT+16 date +%Y-%m-%d_%r`
echo $DATE

---

Related Topics