通过 GNU日期：

  $猫weekof.sh
功能weekof（）
{
    当地周= $1年= $ 16
    当地week_num_of_Jan_1 week_day_of_Jan_1
    当地first_Mon
    当地date_fmt =+％x％A％D％Y
    当地mon·太阳    week_num_of_Jan_1 = $（日期-d $岁的01-01％+ W）
    week_day_of_Jan_1 = $（日期-d $岁的01-01％+ U）    如果（（week_num_of_Jan_1））;然后
        first_Mon = $年01-01
    其他
        first_Mon = $去年01  -  $（（01 +（7  -  week_day_of_Jan_1 + 1）））
    科幻    周一= $（日期-d$ first_Mon + $（（周 -  1））周，$ date_fmt）
    太阳= $（日期-d$ first_Mon + $（（周 -  1））周+ 6天$ date_fmt）
    回声\\$周一\\ -  \\$太阳\\
}weekof $ 1 $ 2
$庆典weekof.sh 12 2012
星期一2012年3月19日 - 太阳2012年3月25日
$庆典weekof.sh 1 2018
星期一2018年1月1日 - 太阳2018年1月7日
$
 

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"

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"
$