How do i get last sunday date from a given date in unix

Question

get last sunday date as output from a given date (not current date) as input

Example input: 08-30-2017 (%m-%d-%Y)

output should be last sunday: 08-27-2017

Answer 1

I hope this below solution can help you:

export day=08-30-2017
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).

Answer 2

Very simple way-

First calculate dayofweek from given date. It will be 1-7. For Monday it's 1 etc ...Saturday 6 and Sunday 7.

Then subtract dayofweek from given date. That's your last Sunday.

$ givenDate="08-30-2017"
$ dayofweek=$(date -j -f '%m-%d-%Y' $givenDate +'%u')
$ date -j -f '%m-%d-%Y' -v-${dayofweek}d $givenDate +%m-%d-%Y
08-27-2017

Answer 3

If you deal with a fixed date format %m-%d-%Y, it should be transformed to %Y-%m-%d format to be processed by date function:

d='08-30-2017'
d=${d##*-}-${d%-*}
lst_sunday=$(date -d "$d -$(date -d $d +%u) days" +"%m-%d-%Y")

echo $lst_sunday
08-27-2017

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+%u - interpreted format specificator, day of week (1..7); 1 is Monday