How to calculate days between two dates

Question

I want a Java program that calculates days between two dates.

1. Type the first date (German notation; with whitespaces: "dd mm yyyy")
2. Type the second date.
3. The program should calculate the number of days between the two dates.

How can I include leap years and summertime?

My code:

import java.util.Calendar;
import java.util.Date;
import java.util.Scanner;

public class NewDateDifference {

    public static void main(String[] args) {

        System.out.print("Insert first date: ");
        Scanner s = new Scanner(System.in);
        String[] eingabe1 = new String[3];

        while (s.hasNext()) {
            int i = 0;
            insert1[i] = s.next();
            if (!s.hasNext()) {
                s.close();
                break;
            }
            i++;
        }

        System.out.print("Insert second date: ");
        Scanner t = new Scanner(System.in);
        String[] insert2 = new String[3];

        while (t.hasNext()) {
            int i = 0;
            insert2[i] = t.next();
            if (!t.hasNext()) {
                t.close();
                break;
            }
            i++;
        }

        Calendar cal = Calendar.getInstance();

Answer 1

You are making some conversions with your Strings that are not necessary. There is a SimpleDateFormat class for it - try this:

SimpleDateFormat myFormat = new SimpleDateFormat("dd MM yyyy");
String inputString1 = "23 01 1997";
String inputString2 = "27 04 1997";

try {
    Date date1 = myFormat.parse(inputString1);
    Date date2 = myFormat.parse(inputString2);
    long diff = date2.getTime() - date1.getTime();
    System.out.println ("Days: " + TimeUnit.DAYS.convert(diff, TimeUnit.MILLISECONDS));
} catch (ParseException e) {
    e.printStackTrace();
}

 Since there have been some discussions regarding the correctness of this code: it does indeed take care of leap years. However, the TimeUnit.DAYS.convert function loses precision since milliseconds are converted to days (see the linked doc for more info). If this is a problem, diff can also be converted by hand:

float days = (diff / (1000*60*60*24));

Note that this is a float value, not necessarily an int.