如何计算日期的差异？

Question

我正在使用java.util.Date 类并想比较Date 对象和当前时间。

输出格式必须是这样的：

... years , ... months , ... days , ...hours , ... minutes , ... seconds.

这是我的代码：

public static void main(String[] args) {
    Calendar calendar = new GregorianCalendar(2022, 1 , 25 , 12 , 20 , 33);
    Date now = new Date(); //Tue Feb 25 11:49:05 IRST 2020
    Date date = calendar.getTime(); //Fri Feb 25 12:20:33 IRST 2022
}

有没有简单的计算方法？

Answer 1

java.time 和 why & how to use it

对于年、月、日的差异，有一个java.time.Period，你可以很容易地得到你想要的：

public static void main(String[] args) {
    LocalDateTime localDateTime = LocalDateTime.of(2022, 1 , 25 , 12 , 20 , 33);
    LocalDateTime now = LocalDateTime.now();

    // get the difference in years, months and days
    Period p = Period.between(now.toLocalDate(), localDateTime.toLocalDate());

    // and print the result(s)
    System.out.println("Difference between " + localDateTime.format(DateTimeFormatter.ISO_LOCAL_DATE_TIME)
            + " and " + now.format(DateTimeFormatter.ISO_LOCAL_DATE_TIME) + " is:\n"
            + p.getYears() + " years, " + p.getMonths() + " months, " + p.getDays() + " days");
}

此代码示例的输出将是（当然取决于当天）：

Difference between 2022-01-25T12:20:33 and 2020-02-25T10:52:43.327 is:
1 years, 11 months, 0 days

您可以使用java.time.Duration，如其他答案之一所示，以获得小时、分钟和秒的额外差异。看这个例子（基本上是上面加时间的计算）：

public static void main(String[] args) {
    LocalDateTime localDateTime = LocalDateTime.of(2022, 1 , 25 , 12 , 20 , 33);
    LocalDateTime now = LocalDateTime.now();

    // get the difference in years, months and days (date related difference)
    Period p = Period.between(now.toLocalDate(), localDateTime.toLocalDate());
    // and the difference in hours, minutes and seconds (time-of-day related difference)
    Duration d = Duration.between(now.toLocalTime(), localDateTime.toLocalTime());
    long totalSeconds = d.getSeconds();
    long hours = totalSeconds / 3600;
    long minutes = (totalSeconds % 3600) / 60;
    long seconds = totalSeconds % 60;

    // and print the result(s)
    System.out.println("Difference between " + localDateTime.format(DateTimeFormatter.ISO_LOCAL_DATE_TIME)
            + " and " + now.format(DateTimeFormatter.ISO_LOCAL_DATE_TIME) + " is:\n"
            + p.getYears() + " years, " + p.getMonths() + " months, " + p.getDays() + " days, "
            + hours + " hours, " + minutes + " minutes, " + seconds + " seconds");
}

输出：

Difference between 2022-01-25T12:20:33 and 2020-02-25T11:25:42.712 is:
1 years, 11 months, 0 days, 0 hours, 54 minutes, 50 seconds

如果您正在接收java.util.Dates 或扩展遗留代码（或者如果您懒得将所有代码更改为使用java.time），您可以使用以下兼容方法：

LocalDateTime fromDate = LocalDateTime.ofInstant(new Date().toInstant(), ZoneId.systemDefault());
Date fromLocalDateTime = Date.from(LocalDateTime.now().toInstant((ZoneOffset.UTC)));

Answer 2

Java 8 Date/Time API 在比较日期方面很灵活。

例子：

LocalDate currentDate = LocalDate.now();
LocalDate previousDate = LocalDate.of(2000, 7, 1);

Period period = Period.between(previousDate, currentDate);

String output = String.format("%s years , %s months , %s days",
                period.getYears(), period.getMonths(), period.getDays());

System.out.println(output);   
> 19 years , 7 months , 24 days

更多示例here。

如果您想在Period 或Duration 之间做出选择，This 可能会对您有所帮助。

Answer 3

使用 java Instant

Instant start, end;
Duration duration = Duration.between(start, stop);
long hours = dur.toHours();
long minutes = dur.toMinutes();