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

I think this actually answers what was requested:

date -d "next monday - 7 days"

If you want the monday previous to a given date, like a start of week function, there's a little more to it. That's what I was looking for when I searched up this question, so I'll add my answer here.

I tried to take a "reusable tool" approach.

My monday script:

#!/bin/bash
date=${1:-now}
shift
dow=`date +%u -d "$date"`
dow0=$[ $dow - 1 ]
date -d "$date -$dow0 days" "$@"

How it works:

• get date -d argument from first command line argument, e.g. "2020-01-01 -2 weeks"
• additional args are passed on to the final date command, e.g. to format with +%F
• get the day of week with %u (1..7); 1 is Monday
• subtract 1 as we want the range 0 to 6.
• show the date for the given date, minus the "day of the week" offset

A similar sunday script, which turns out to be a little simpler:

#!/bin/bash
date=${1:-now}
shift
dow=`date +%w -d "$date"`
date -d "$date -$dow days" "$@"

Usage:

# monday 
Mon Aug 29 02:28:45 PM AEST 2022
# sunday now +%F
2022-08-28
# monday '2022-12-01' +%F
2022-11-28