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Help, I need to get the last date of previous month


 
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# 1  
Old 02-25-2008
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.
# 2  
Old 02-25-2008
continuing in the style of your script try using code like:

Code:
/bin/cal $prev_month $cur_year | grep -v "^$" | sed -n  '1p;$p' | tr "\n" " "  | awk '{print substr($1,1,3),$NF,$2}'

HTH
# 3  
Old 02-25-2008
Tytalus,

Thanks for the quick reply, your code is giving me a format that I needed but It gave me the current month. I looking for a last date of the previous month.
Please advice.

Thank you.
# 4  
Old 02-25-2008
I got it - sorry

Thank you very much for your help.
# 5  
Old 02-25-2008
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

# 6  
Old 02-28-2008
Another question, how do I get the first of a previous month?
I'm not familia with sed.

Thank you.
# 7  
Old 02-28-2008
Code:
/bin/cal $month $year| grep -v "^$" | sed -n  '1p;3p' | tr "\n" " "  | awk '{print substr($1,1,3),$3,$2}'

Should work ;-)

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