How to determine number of week of the month

Question

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.

Answer 1

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

Answer 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) ))

Answer 3

Try this:

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

Answer 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.

Answer 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