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.
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How do you define the week of the month? The day number divided by 7 and rounded down? Number of Sundays/Mondays/whatever which have passed?– mgorvenApr 27, 2012 at 5:23
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lets say from one to 7 day its 1rst week, from 7 to 14 second week, from 14 to 21 third and from 21 to end of the month its fourth week. I know that simple bash script with date nad if is nice solution for this but im wondering if i can di that with one command without doing a script.– B14D3Apr 27, 2012 at 5:28
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@B14D3 your "week of month" definition is pretty coarse -- what specifically are you using this for?– voretaq7Apr 27, 2012 at 5:47
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1For nothing I was reading man for date and that came to my mind (Sometimes I have so silly thoughts). So what will be a better definition ?– B14D3Apr 27, 2012 at 5:54
5 Answers
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
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That work fine, but I dont understand why this arithmetic returns integer ?– B14D3Apr 27, 2012 at 5:50
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5Sorry I dont know how to format the comment to look nicer. Anyway, great idea to implement, however if the date is 08 or 09, it will cause error: -bash: (09: value too great for base (error token is "09"). Thats is because if numerical value starts with 0, it will be intepreted as octal number by C language, so 08 09 are invalid. For me, the workaround is to change from %d to %e, %e omits the leading 0: echo $((($(date +%e)-1)/7+1)) Sep 9, 2013 at 3:55
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) ))
Try this:
d=`date +%d` ; m=`date +%m` ; y=`date +%Y` ; cal $m $y | sed -n "3,$ p" | sed -n "/$d/{=;q;}"
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.
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