Java Programming logic for parking fee?

Question

I Have a problem with this logic in the car parking the program is i need to use calendar from the api to get the hour in the time in and time out. I have a problem with the overnight parking because i use the hour in the system time if i time In in 20 o clock and i want to time out in like 5 oclock the computation is ruined can you give me any suggestions on how to do this

1st Class

public int price()
{
int T = Out - In;

if(Tl == 0)
return 0;
else if(T < 3)
return ITime;
else
T -= 3;
return (5 * T) + 15;
}

2nd Class
Calendar cal = Calendar.getInstance();
c.setcar(cal.getTime().getHours());
if((c.In() >=1 ) && c.Out() == 0))
JOptionPane.showMessageDialog(null, " OverNightParking: Your Fee is 200 Pesos");
else if(c.price() >= 60)
JOptionPane.showMessageDialog(null, " Your Parking Fee is 60 ");
else if(c.price() == 0)
JOptionPane.showMessageDialog(null, " Your Parking Fee is free ");
else
JOptionPane.showMessageDialog(null, "Your Parking Fee is " + car1.price() + "$");

Answer 1

Don't forget you can use the AM and PM fields to check if it's PM or AM. It will be 0 if Am and 1 if PM if I recall.



So 20 o clock would have an AM_PM value of 1 while 5 o clock would have an AM_PM value of 0.

Answer 2

When we 'math' look for smallest unit. With time it is milliseconds.



import java.util.Calendar;





public class ParkTktDiff {



public static void main( String [] args ) {

Calendar today = Calendar.getInstance();

Calendar later = Calendar.getInstance();

long ticToc = 60*60*1000; // 1 hour

ticToc += 10*60*1000; // ten min. more

ticToc += later.getTimeInMillis(); // get the start of time

later.setTimeInMillis(ticToc); // set time forward



long diff = later.getTimeInMillis() - today.getTimeInMillis();



System.out.printf("DooWad Parking: %2d hours : %2d mins%n",

diff / (60*60*1000),

(diff / (60*1000)) - 60);



}

}

//.====================

long diffSeconds = diff / 1000;

long diffMinutes = diff / (60 * 1000);

long diffHours = diff / (60 * 60 * 1000);

long diffDays = diff / (24 * 60 * 60 * 1000);

Answer 3

If the calendar object can also give you the date you can use that to check if the date has changed and use it to calculate the fee.

Answer 4

f CV mve ,dzx