Run script problem

I hv a script as below , it work fine to run at unix box , but fail to
run at linux , could suggest how to modify it to make it work fine
at linux , very thx if anyone can help.

my script is as below , could suggest how can I modify it to make it
work ? thx


#!/bin/sh

PATH=$PATH:/usr/bin:/usr/sbin:/usr/contrib/bin
TIMEOUT=40
EXCEPTIONLIST=~ckyoung/exception.lst
touch $EXCEPTIONLIST

kill_pid()
{
PID=$1
kill -9 $PID
}

echo "User:Activity:PID:Status"
who -u | while read line
do
time=$(echo $line | awk '{print $6}')
pid=$(echo $line | awk '{print $7}')
user=$(echo $line | awk '{print $1}')

grep -q "^${user}:" $EXCEPTIONLIST
if [ $? != 0 ]
then
if [ "$time" = "old" ]
then
(( time = $TIMEOUT + 1 ))
else
echo $time |grep -q "\."
if [ $? != 0 ]
then
time=$(echo $time|awk '{FS=":";print $1*60+$2}')
else
time=0
fi
fi
fi
#echo $time
if [ $time -gt $TIMEOUT ]
then
echo "$user:$time:$pid:Killed"

# Uncomment to do kill that PID
kill_pid $pid
else
echo "$user:$time:$pid:Untouched"
fi
done

the error is
../kill_idle: line 38: [: 00:03: integer expression expected , where
line 38 is "if [ $time -gt $TIMEOUT ]" ,

On 2005-01-05, (E-Mail Removed) <(E-Mail Removed)> wrote:
> the error is
> ./kill_idle: line 38: [: 00:03: integer expression expected , where
> line 38 is "if [ $time -gt $TIMEOUT ]" ,

Yes, well, obviously "00:03" isn't recognized as a number. You'll have
to fix the way you get '$time'.

Davide

--
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