我们可以利用LocalDate和ChronoUnit java库，下面的代码工作正常。 Date的格式为yyyy-MM-dd。

import java.time.LocalDate;
import java.time.temporal.ChronoUnit;
import java.util.*;
class Solution {
    public int daysBetweenDates(String date1, String date2) {
        LocalDate dt1 = LocalDate.parse(date1);
        LocalDate dt2= LocalDate.parse(date2);

        long diffDays = ChronoUnit.DAYS.between(dt1, dt2);

        return Math.abs((int)diffDays);
    }
}

String dateStart = "01/14/2015 08:29:58";
String dateStop = "01/15/2015 11:31:48";

//HH converts hour in 24 hours format (0-23), day calculation
SimpleDateFormat format = new SimpleDateFormat("MM/dd/yyyy HH:mm:ss");

Date d1 = null;
Date d2 = null;

d1 = format.parse(dateStart);
d2 = format.parse(dateStop);

//in milliseconds
long diff = d2.getTime() - d1.getTime();

long diffSeconds = diff / 1000 % 60;
long diffMinutes = diff / (60 * 1000) % 60;
long diffHours = diff / (60 * 60 * 1000) % 24;
long diffDays = diff / (24 * 60 * 60 * 1000);

System.out.print(diffDays + " days, ");
System.out.print(diffHours + " hours, ");
System.out.print(diffMinutes + " minutes, ");
System.out.print(diffSeconds + " seconds.");

更新:2013年的原始答案现在已经过时，因为一些类已经被替换了。新的方法是使用新的java。时间类。

DateTimeFormatter dtf = DateTimeFormatter.ofPattern("dd MM yyyy");
String inputString1 = "23 01 1997";
String inputString2 = "27 04 1997";

try {
    LocalDateTime date1 = LocalDate.parse(inputString1, dtf);
    LocalDateTime date2 = LocalDate.parse(inputString2, dtf);
    long daysBetween = Duration.between(date1, date2).toDays();
    System.out.println ("Days: " + daysBetween);
} catch (ParseException e) {
    e.printStackTrace();
}

注意，这个解决方案给出的是实际的24小时天数，而不是日历天数。对于后者，请使用

long daysBetween = ChronoUnit.DAYS.between(date1, date2)

原始答案(Java 8已过时)

您正在使用字符串进行一些不必要的转换。有一个SimpleDateFormat类-试试这个:

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();
}

编辑:由于有一些关于这段代码正确性的讨论:它确实考虑到了闰年。然而，TimeUnit.DAYS.convert函数失去了精度，因为毫秒被转换为天(更多信息请参阅链接文档)。如果这是一个问题，diff也可以手动转换:

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

注意，这是一个浮点值，不一定是整型。

当夏令时到来时，大多数/所有答案都给我们带来了问题。下面是我们针对所有日期的工作解决方案，不使用JodaTime。它利用日历对象:

public static int daysBetween(Calendar day1, Calendar day2){
    Calendar dayOne = (Calendar) day1.clone(),
            dayTwo = (Calendar) day2.clone();

    if (dayOne.get(Calendar.YEAR) == dayTwo.get(Calendar.YEAR)) {
        return Math.abs(dayOne.get(Calendar.DAY_OF_YEAR) - dayTwo.get(Calendar.DAY_OF_YEAR));
    } else {
        if (dayTwo.get(Calendar.YEAR) > dayOne.get(Calendar.YEAR)) {
            //swap them
            Calendar temp = dayOne;
            dayOne = dayTwo;
            dayTwo = temp;
        }
        int extraDays = 0;

        int dayOneOriginalYearDays = dayOne.get(Calendar.DAY_OF_YEAR);

        while (dayOne.get(Calendar.YEAR) > dayTwo.get(Calendar.YEAR)) {
            dayOne.add(Calendar.YEAR, -1);
            // getActualMaximum() important for leap years
            extraDays += dayOne.getActualMaximum(Calendar.DAY_OF_YEAR);
        }

        return extraDays - dayTwo.get(Calendar.DAY_OF_YEAR) + dayOneOriginalYearDays ;
    }
}

想要得到只是天(没有时间)，你可以使用ChronoUnit

ChronoUnit.DAYS.between(date1.toLocalDate(), date2.toLocalDate());

Use:

public int getDifferenceDays(Date d1, Date d2) {
    int daysdiff = 0;
    long diff = d2.getTime() - d1.getTime();
    long diffDays = diff / (24 * 60 * 60 * 1000) + 1;
    daysdiff = (int) diffDays;
    return daysdiff;
}