UNIX date: How to convert week number (date +%W) to a date range (Mon-Sun)?

Question

I have list of week numbers extracted from huge log file, they were extracted using syntax:

$ date --date="Wed Mar 20 10:19:56 2012" +%W;
12

I want to create a simple bash function which can convert these week numbers to a date range. I suppose function should accept 2 arguments: $number and $year, example:

$ week() { ......... }
$ number=12; year=2012
$ week $number $year
"Mon Mar 19 2012" - "Sun Mar 25 2012"

Answer 1

With GNU date:

$ cat weekof.sh
function weekof()
{
    local week=$1 year=$2
    local week_num_of_Jan_1 week_day_of_Jan_1
    local first_Mon
    local date_fmt="+%a %b %d %Y"
    local mon sun

    week_num_of_Jan_1=$(date -d $year-01-01 +%W)
    week_day_of_Jan_1=$(date -d $year-01-01 +%u)

    if ((week_num_of_Jan_1)); then
        first_Mon=$year-01-01
    else
        first_Mon=$year-01-$((01 + (7 - week_day_of_Jan_1 + 1) ))
    fi

    mon=$(date -d "$first_Mon +$((week - 1)) week" "$date_fmt")
    sun=$(date -d "$first_Mon +$((week - 1)) week + 6 day" "$date_fmt")
    echo "\"$mon\" - \"$sun\""
}

weekof $1 $2
$ bash weekof.sh 12 2012
"Mon Mar 19 2012" - "Sun Mar 25 2012"
$ bash weekof.sh 1 2018
"Mon Jan 01 2018" - "Sun Jan 07 2018"
$

---

NOTE:

As the OP mentions, the week number is got by date +%W. According to GNU date's manual:

%W: week number of year, with Monday as first day of week (00..53)

So:

1. Each week starts from Mon.
2. If Jan 1 is Mon, then the first week will be week #1.
3. If Jan 1 is not Mon, then the first few days will be week #0 and the week #1 starts from the first Mon.