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This is a little trickier than finding the last Sunday, because you know the last Sunday is in the first position of the last line. The trick is to use the NF less than or equal to 7 so it picks up all the lines then grep out any empty lines.
There is 1 alternative - vote for the best!
If your locale has Monday as the first day of the week, like mine in the UK, change the two $7 into $6
Choosing your year and month. You only need the gnu date command and bash. desiredDay of the week is (1..7); 1 is Monday.
If you want desiredDay of week (0..6); 0 is Sunday
desiredDay=6; year=2012; month=5; n=0; while [ $(date -d "$year-$((month+1))-1 - $n day" "+%w") -ne $desiredDay ]; do n=$((n+1)); done; date -d "$year-$((month+1))-1 - $n day" "+%x"
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Uh, not sure. It's May here and I get last day of the current month.
31 using
cal 04 2012 | awk '{ $7 && X=$7 } END { print X }
@habitual: that's what the '04' parameter is: April. Replace it with 05 for May.
flatcap: But what if I WANT the last day of 04?
I think cal now supports knowing the first day of the week depending on your _NL_TIME_WEEK_1STDAY. Does your output of `cal 05 2012` equal:
May 2012
Su Mo Tu We Th Fr Sa
1 2 3 4 5
6 7 8 9 10 11 12
13 14 15 16 17 18 19
20 21 22 23 24 25 26
27 28 29 30 31