bash脚本，如果没有提供，请询问arg

我已经创建了一个脚本，该脚本将检查您提供的用户是否已登录，并在登录时显示会话的持续时间。我现在需要做的是，如果在发出命令时没有提供参数(用户名(，则要求提供一个参数，并获得与您提供的参数相同的结果。

这是我所拥有的：

name=$(cat /etc/passwd | grep $1 | cut -d':' -f5 | tr ':' ' ' | sed 's/,//' | sed 's/^([^ ]*) ([^ ]*)/2 1/' | sort -t' ' -k3,3)
terminal=$(who | grep $1 | cut -d' ' -f3)
loginHour=$(who | grep $1 | cut -c30-31)
loginMin=$(who | grep $1 | cut -c33-34)
loginMins=$((loginHour * 60 + loginMin))
nowHour=$(date +%R | cut -c1-2)
nowMin=$(date +%R | cut -c4-5)
nowMins=$((nowHour * 60 + nowMin))
totalMins=$((nowMins - loginMins))
hoursOn=$((totalMins / 60))
minsOn=$((totalMins % 60))
clear
echo
if [[ $# -eq 1 ]] ; then
    grep -q $1 /etc/passwd
    if grep -q $1 /etc/passwd ; then
        clear
        echo
        if who | grep $1 > /dev/null ; then
            echo "$name" is currently logged on to terminal "$terminal" and has been for "$hoursOn" hour"(s)" and "$minsOn" minute"(s)".
            echo
            exit 0
        else
            echo "$name" is NOT currently logged on.
            echo
            exit 1
        fi
    else
        echo The user you entered is not a valid user on this system.
        echo
        exit 2
    fi
fi

我以前尝试过，但没有达到预期的结果，所以我把它去掉了。

if [[ $# -eq 0 ]]
then
    read -p "Enter Name: " username
else
    username=$1
fi

# : does nothing it just forces the evaluation
: ${1:?"Need to provide name to script"}