Print Date for the Monday of the Current Week (In Bash)

Print date for the monday of the current week (in bash)

#monday
date -dmonday +%Y%m%d

#last monday
date -dlast-monday +%Y%m%d

#next monday
date -dnext-monday +%Y%m%d

#two mondays from now
date -d'monday+14 days' +%Y%m%d

#two mondays ago
date -d'monday-14 days' +%Y%m%d

#although, if you fancy yourself an Abraham Lincolin
date -d'monday-fortnight ago' +%Y%m%d #2 weeks ago
date -d'monday+fortnight' +%Y%m%d #2 weeks from now

#Monday Next Year
date -d'52+monday' +%Y%m%d

#However, Monday Last Year
date -d'52-monday' +%Y%m%d  #DOES NOT  WORK

#you can try a day other than monday
#and format this differently.

if a range is what your after you may need to do a few things

#Tuesday to Sunday
#since today is monday, I'll use Tuesday
echo `date -dtuesday +%Y%m%d-``date -dnext-sunday +%Y%m%d`

which would output:

20110628-20110703

More on Dates

note this only works on GNU date

I have read that:

Solaris version of date, which unable
to support -d can be resolve with
replacing sunfreeware.com version of
date

Get monday of the currentweek with linux date

date -d "next-monday - 7 days"

date -d "$(($(date +%u) -1)) days ago"

# today
date  +%Y%m%d
20220524

# Monday
$ date -d "next-monday - 7 days" +%Y%m%d
20220523
# or
$ date -d "$(($(date +%u) -1)) days ago" +%Y%m%d
20220523

Bash- getting week start from date

Try:

$ date -d "$(date -d yesterday +%u) days ago"
Mon Aug 21 18:18:54 PDT 2017

How it works:

• date -d yesterday +%u gets yesterday's day of the week (1=Monday, 7=Sunday).

• date -d "$(date -d yesterday +%u) days ago" returns the date for enough days ago to get the monday before yesterday.

  For example, since today is Sunday, yesterday's day of week is 6 (Saturday). 6 days ago is last monday.

  If today was Monday, yesterday's day of the week would be Sunday which is 7 and "7 days ago" would be the monday before today.

Get Monday and Sunday etc.. for a week for any date as parameter in Unix

Try this:

export day=2013-10-01
date -d "$day -$(date -d $day +%w) days"

This will always print the Sunday before the given date (or the date itself).

date -d "$day -$(date -d $day +%u) days"

This will always print the Sunday before the given date (and never the date itself).

For Mondays you need to add + 1 day:

date -d "$day -$(date -d $day +%u) days + 1 day"

You should also consider what Monday or Sunday you want to get (this wasn't quite clear) for which date. This also depends on whether you consider the Sunday the first or the last day of the week.

How do I get the current week start date and end date in ksh

If you have Gnu or recent versions of date, this works.

date -dlast-sunday +%d%m%y

date -dnext-saturday +%d%m%y

Use Perl on Solaris because date command does not support this.

perl -MPOSIX -e '@DateComponents = localtime; print strftime "%d%m%y", localtime time - 86400*(($DateComponents[6]))' 

perl -MPOSIX -e '@DateComponents = localtime; print strftime "%d%m%y", localtime time + 86400*(6-($DateComponents[6]))'

BASH: print the next monday of date

If GNU date (or any other date) accepts some variation of "monday after $DAYCHECK", I haven't seen it. I think you have to do the math.

$ day_of_week=$( date +%u --date $DAYCHECK)   # 1 = Monday, ... 7 = Sunday
$ date --date "$DAYCHECK +$((8-day_of_week)) days"

If DAYCHECK is already a Monday, you'll get the following Monday. If you want DAYCHECK instead, you'll need to handle it separately:

$ if (( day_of_week == 1 )); then
> date --date "$DAYCHECK"
> else
> date --date "$DAYCHECK +$((8-day_of_week)) days"
> fi

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