How to get the latest date for a specific day of the week in Bash [duplicate] - bash

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How to find out the date of the last Saturday in Linux shell script or python?
(3 answers)
Closed 6 years ago.
Q: How can the latest date for a given_day_of_the_week?
Example: For example if today is the 2016-04-21, I just want to get the date for a given_day_of_the_week for example Monday which would be the 2016-04-18. This date will be still returned if the code was run tomorrow, up until the day before the following given_day_of_the_week.
Explanation: The code should return the latest given_day_of_the_week's (Monday) date until the Sunday 2016-04-24, then the same bit of code running will return the 2016-04-25 next week.

Answer 1

Requires GNU date:
today_dow=$(date +%w)
days=(Sunday Monday Tuesday Wednesday Thursday Friday Saturday)
for ((dow=0; dow<7; dow++)); do
if ((dow < today_dow)); then
date -d "last ${days[dow]}"
else
date -d "${days[dow]}"
fi
done
Sun Apr 17 00:00:00 EDT 2016
Mon Apr 18 00:00:00 EDT 2016
Tue Apr 19 00:00:00 EDT 2016
Wed Apr 20 00:00:00 EDT 2016
Thu Apr 21 00:00:00 EDT 2016
Fri Apr 22 00:00:00 EDT 2016
Sat Apr 23 00:00:00 EDT 2016
So we can do:
given_day_of_the_week() {
local days=(Sunday Monday Tuesday Wednesday Thursday Friday Saturday)
local today_dow=$(date +%w)
local dow datestr
for ((dow=0; dow<7; dow++)); do
if [[ "${days[dow],,}" == "${1,,}" ]]; then
if ((dow < today_dow)); then
datestr="last ${days[dow]}"
else
datestr="${days[dow]}"
fi
date -d "$datestr" "+%F"
fi
done
}
Resulting in:
$ given_day_of_the_week tuesday
2016-04-19
$ given_day_of_the_week friday
2016-04-22
Hardcoding the weekday names like that will give you problems if you're in a different locale
Responding to #ryenus's comment:
$ given_day_of_the_week() {
local -A days=([sunday]=0 [monday]=1 [tuesday]=2 [wednesday]=3 [thursday]=4 [friday]=5 [saturday]=6)
local today_dow=$(date +%w)
local datestr=${1,,}
local dow=${days[$datestr]}
[[ -z "$dow" ]] && { echo "error: unknown day: '$1'" >&2; return 1; }
(( dow < today_dow )) && datestr="last $datestr"
date -d "$datestr" "+%F"
}
$ given_day_of_the_week friday
2016-04-22
$ given_day_of_the_week monday
2016-04-18
$ given_day_of_the_week FOO
error: unknown day: 'FOO'

Answer 2

Inspired by #glenn-jackman's answer, but simpler:
given_day_of_the_week=$1
f='%A %Y-%m-%d'
tomorrow=`date -d 'tomorrow' +"$f"`
today=`date -d 'today' +"$f"`
last5=`seq 1 5 | xargs -I{} date -d '{} day ago' +"$f"`
echo -e "$tomorrow\n$today\n$last5" |
grep -i $given_day_of_the_week |
cut -d' ' -f2
Note: this answer does give tomorrow's date for Monday if used on a Sunday. It also gives tomorrow's date for Tuesday if used on a Monday. Not 100% sure if that's what you had in mind.
If instead the idea is to always go with the current Sunday - Saturday, this will do it:
given_day_of_the_week=$1
f='%A %Y-%m-%d'
[ `date +%A` = 'Sunday' ] && first=`date -d 'today' +"$f"` ||
first=`date -d 'last Sunday' +"$f"`
rest=`seq 1 6 | xargs -I{} date -d "$first +{} day" +"$f"`
echo -e "$first\n$rest" |
grep -i $given_day_of_the_week |
cut -d' ' -f2

Answer 3

# requires bash 4.x and GNU date
last_kday() {
local kday=$1
local -A numbers=([sunday]=0 [monday]=1 [tuesday]=2 [wednesday]=3
[thursday]=4 [friday]=5 [saturday]=6)
if [[ $kday == *day ]]; then
kday=${numbers[${kday,,}]}
elif [[ $kday != [0-6] ]]; then
echo >&2 "Usage: last_kday weekday"
return 1
fi
local today=$(date +%w)
local days_ago=$(( today - kday ))
if (( days_ago < 0 )); then let days_ago+=7; fi
date -d "$days_ago days ago" +%F
}