Shell Programming and Scripting

Hi, i;m beginner of Unix, i trying to use crontab to zip my log file automatically, below is my coding, some of the statement i don't know whether is correct or not. Pls help

year=`date '+%Y'`
month=`date '+%m'`
day=`date '+%d'`
day=`expr $day - 1`

case $month in
1 | 3 | 5 | 7 | 8 | 9 | 10 | 12);;
if [ $day = 0 ]; then
day=32
day1=$day-1
fi

4 | 6 | 11);;
if [ $day = 0 ]; then
day=31
day1=$day-1'
fi

2);;
if [ $month Mod 4 =0]; then
if[ $day = 0 ]; then
day = 30
day1 =%day-1
fi
else
if[ $day = 0 ]; then
day = 29
day1 =$day-1'
fi
fi

*) echo error: too many arguments 1>&2 ;;
esac

DATE=$year$month$day1
mv $HOME/*$DATE*log $HOME/archive
cd $HOME/archive
gzip *

Please describe any bugs you get and ask specific questions. Nobody here will do the testing guy for you.

Why don't you simply use logratate which does exactly the same thing?

Use the code tags it will help in understanding the problem clearly with identation.
in tool box the hash button is for the code tags.

Thanks

Buddy,

The code seems to be correct. But it needs certain modifications at some places. Please find below the corrected code:

year=`date '+%Y'`
month=`date '+%m'`
day=`date '+%d'`
day=`expr $day - 1`

case $month in
1 | 3 | 5 | 7 | 8 | 9 | 10 | 12);;
if [ $day = 0 ]; then
day=32
day1=`expr $day-1`
fi

4 | 6 | 11);;
if [ $day = 0 ]; then
day=31
day1=`expr $day-1`
fi

2);;
if [ `expr $month % 4` -eq 0]; then
if[ $day = 0 ]; then
day = 30
day1 =`expr $day-1`
fi
else
if[ $day = 0 ]; then
day = 29
day1 =`expr $day-1`
fi
fi

*) echo error: too many arguments 1>&2 ;;
esac

DATE=$year$month$day1
mv $HOME/*$DATE*log $HOME/archive
cd $HOME/archive
gzip *


Probably this would work. Just check out the differences with your code. Revert back in case of any issue.

Thanks so much guys...i'll try it out.