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B14D3
Asked: 2012-04-27 21:21:25 +0800 CST

How to determine number of week of the month

  • 772

On the Linux CLI, is there a way to get the number of the week of the month? Maybe there is another way to get this with one simple (like date) command? Let's say that day 1 to 7 is the first week, day 8 to 14 is the second week, and so on.

linux date

5 Answers

  1. Best Answer

    The date command can't do this internally, so you need some external arithmetic. 

    echo $((($(date +%-d)-1)/7+1))

    Edit: Added a minus sign between the % and the d

    • 23

  2. You can use this:

    Monday First week day

    WEEKNUMBER=$(( 1 + $(date +%V) - $(date -d "$(date -d "-$(($(date +%d)-1)) days")" +%V) ))

    Sunday Firs week daty

    WEEKNUMBER=$(( 1 + $(date +%U) - $(date -d "$(date -d "-$(($(date +%d)-1)) days")" +%U) ))
    • 2

  3. Try this:

    d=`date +%d` ; m=`date +%m` ; y=`date +%Y` ; cal $m $y | sed -n "3,$ p" | sed -n "/$d/{=;q;}"
    • 1

  4. simplifying Victor Sanchez's solution:

    expr 1 + $(date +%V) - $(date +%V -d $(date +%Y-%m-01))

    replace %V with %U if you want weeks starting on Sunday.

    btw: had to use expr instead of $((...)) because the later doesn't seem to like numbers with leading zeroes.

    • 1

  5. If you accept external tools in your quest, try dateutils. It's got the notion of occurrence-within-month dates, i.e. 27 Apr 2012 is the 4th Fri in Apr 2012, which just coincides with your week definition. To get that number use:

    dconv 2012-04-27 -f %c
=>
  04

    %c (count) is the format specifier for the occurrence-within the month. Or to be even cooler try

    dconv today -f '%cth %a in %b %Y'
=>
  1st Wed in Sep 2012
    • 0