bash – UNIX日期:如何将周数转换为日期范围(Mon-Sun)?

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我有从大日志文件提取的周数列表,它们是使用语法提取的:
  1. $date --date="Wed Mar 20 10:19:56 2012" +%W;
  2. 12

我想创建一个简单的bash函数,可以将这些周数转换为日期范围.我想函数应该接受2个参数:$number和$year,例如:

  1. $week() { ......... }
  2. $number=12; year=2012
  3. $week $number $year
  4. "Mon Mar 19 2012" - "Sun Mar 25 2012"
使用GNU日期:
  1. $cat weekof.sh
  2. function weekof()
  3. {
  4. local week=$1 year=$2
  5. local week_num_of_Jan_1 week_day_of_Jan_1
  6. local first_Mon
  7. local date_fmt="+%a %b %d %Y"
  8. local mon sun
  9.  
  10. week_num_of_Jan_1=$(date -d $year-01-01 +%W)
  11. week_day_of_Jan_1=$(date -d $year-01-01 +%u)
  12.  
  13. if ((week_num_of_Jan_1)); then
  14. first_Mon=$year-01-01
  15. else
  16. first_Mon=$year-01-$((01 + (7 - week_day_of_Jan_1 + 1) ))
  17. fi
  18.  
  19. mon=$(date -d "$first_Mon +$((week - 1)) week" "$date_fmt")
  20. sun=$(date -d "$first_Mon +$((week - 1)) week + 6 day" "$date_fmt")
  21. echo "\"$mon\" - \"$sun\""
  22. }
  23.  
  24. weekof $1 $2
  25. $bash weekof.sh 12 2012
  26. "Mon Mar 19 2012" - "Sun Mar 25 2012"
  27. $bash weekof.sh 1 2018
  28. "Mon Jan 01 2018" - "Sun Jan 07 2018"
  29. $

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