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02-25-2008
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Help, I need to get the last date of previous month
Hi, I'm new with Unix, I'm trying to get a last day of previous month with this format: %b %d %Y (example: Feb 25 2008).
Here is what I have so far. #!/bin/ksh cur_month=`date +%m` cur_year=`date +%Y` prev_month=$(($cur_month-1)) # Check to see if this is January if [ $prev_month -lt 1 ] then prev_year=$(($cur_year-1)) prev_month=12 LastDayOfMonth=`/bin/cal $prev_month $prev_year | grep -v "^$" | tail -1 | awk '{print $NF}'` else LastDayOfMonth=`/bin/cal $prev_month $cur_year | grep -v "^$" | tail -1 | awk '{print $NF}'` fi ============================================== I can get the last date of the previous month but I don't know how to put it back into the format that I wanted. Please help. Thank you. |
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02-25-2008
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Code:
# cat lastday.ksh
#!/bin/ksh
cur_month=`date +%m`
cur_year=`date +%Y`
prev_month=$(($cur_month-1))
# Check to see if this is January
if [ $prev_month -lt 1 ]
then
prev_year=$(($cur_year-1))
prev_month=12
LastDayOfMonth=`/bin/cal $prev_month $cur_year | grep -v "^$" | sed -n '1p;$p' | tr "\n" " " | awk '{print substr($1,1,3),$NF,$2}'`
else
LastDayOfMonth=`/bin/cal $prev_month $cur_year | grep -v "^$" | sed -n '1p;$p' | tr "\n" " " | awk '{print substr($1,1,3),$NF,$2}'`
fi
echo $LastDayOfMonth
# ./lastday.ksh
Jan 31 2008
# date
Mon Feb 25 16:23:48 GMT 2008
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02-28-2008
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Code:
/bin/cal $month $year| grep -v "^$" | sed -n '1p;3p' | tr "\n" " " | awk '{print substr($1,1,3),$3,$2}'
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