Question

1. Write a shell script called userMonitor.sh that checks each minute and reports on who logs in and who logs out. Please ignore multiple logins by the same user.
2. Run in unix with screenshot

Answer 1

Hi,

#!/usr/bin/ksh

typeset -i SINCE=${1:-1}

date '+%Y +%m %d %H %M' | read year month day hour min

(( min -= SINCE ))

(( min < 0 )) && { (( min += 60 )) ; (( hour -= 1 )) }

(( hour < 0 )) && { (( hour += 24 )) ; (( day -= 1 )) }

(( day <= 0 )) && {

(( month -= 1 ))

(( month <= 0 )) && { (( month=12 )) ; (( year -= 1 )) }

day=$(cal $month $year | tr '\n' ' ' | awk '{print $NF}')

}

ts5=$(printf "%02d%02d%02d%02d" $month $day $hour $min)   

LANG=C who -s |

awk -v ts5="$ts5" -F '[ \t:]*' '

function to_timestamp(month, day, hour, min) {

m = index("_,jan,feb,mar,apr,may,jun,jul,aug,sep,oct,nov,dec,", "," tolower(month) ",")/4;

return sprintf("%02d%02d%02d%02d", m, day, hour, min);

}

to_timestamp($3, $4, $5, $6) >= ts5

'

$