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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"
There are 2 alternatives - 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
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.
If you can do better, submit your command here.
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