Syntax in ksh


Hi,

 I get the following error :  syntax error at line 10 : `else' unmatched

My code looks like. It returns the last day of the month so its a 30,31 for most or 28 or 29 for Feb depending upon whether its a leap year. Any ideas?

#! /usr/bin/ksh
 
y=`date +"%Y"`
m=`date +"%m"`
d=`date +"%d"`
 
if [[ $m == 01 || $m == 03 || $m == 05 || $m == 07 || $m == 08 || $m == 10 ||
$m == 12 ]]
then endD=31;
else if [[ $m == 02 ]]
 then
  leapyr=`expr $y%4`;
  if [[ $leapyr -eq 0 ]]
   then
    endD=29;
   else
    endD=28;
  fi
 else
endD=30;
fi
 
echo $y
echo $m
echo $endD
LuckyLucksAsked:
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TintinCommented:
Much quicker and easier to do
#!/bin/ksh
cal|awk 'END{print DAY};{if (NF<1) {next};DAY=$NF}'

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0
LuckyLucksAuthor Commented:
Its for the last month's end date
0
TintinCommented:
Your original code was trying to get the last date from the current month.

What Unix/Linux system are you trying to do this on?
0
LuckyLucksAuthor Commented:
unix AIX 5.3
Sorry my wording was off, am looking for the last day of previous month
0
TintinCommented:
So long as you have GNU date on your system (you can easily install it if you need to)

The following will output the last day of the previous month.  

date -d "$(date -d "$(date +%Y-%m-%d) -1 month" +%Y-%m-01) +1 month -1 day" +%d
0

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