如何在Java中以分钟为单位获取两个日期之间的差异

Question

我有一个如下所示的日期列表：

I/System.out: TAG: Mon Jun 20 15:36:00 GMT+02:00 2022
I/System.out: TAG: Mon Jun 20 15:51:55 GMT+02:00 2022
I/System.out: TAG: Mon Jun 20 15:52:14 GMT+02:00 2022
I/System.out: TAG: Mon Jun 20 16:24:01 GMT+02:00 2022
I/System.out: TAG: Mon Jun 20 18:39:07 GMT+02:00 2022
I/System.out: TAG: Mon Jun 20 18:49:02 GMT+02:00 2022
I/System.out: TAG: Mon Jun 20 20:24:26 GMT+02:00 2022
I/System.out: TAG: Mon Jun 20 20:52:23 GMT+02:00 2022
I/System.out: TAG: Mon Jun 20 22:41:50 GMT+02:00 2022
I/System.out: TAG: Mon Jun 20 22:55:54 GMT+02:00 2022

如何计算从 i 元素中减去每个 i+1 元素（例如：标签：2022 年 6 月 20 日星期一 15:51:55 GMT+02:00 2022 年 - 标签：2022 年 6 月 20 日星期一 15:36:00 GMT+02:00 2022) 相差 21 分钟。

我尝试了这样的事情：

 for(int i=0;i<list.size()-1;i++){
                     parked += list.get(i+1).getTime() - list.get(i).getTime();
                    }
System.out.println("parked time:  "+parked / 60 000);

我得到 434 分钟而不是 98 分钟......有人可以帮忙吗？ 致谢

Answer 1

通过查看您的问题，我发现了能够解决您问题的其他帖子。 作为解决方案提供的代码能够将 Date 转换为简单的可读值。

计算两个 Java 日期实例之间的差异

/**
 * Get a diff between two dates
 * @param date1 the oldest date
 * @param date2 the newest date
 * @param timeUnit the unit in which you want the diff
 * @return the diff value, in the provided unit
 */
public static long getDateDiff(Date date1, Date date2, TimeUnit timeUnit) {
    long diffInMillies = date2.getTime() - date1.getTime();
    return timeUnit.convert(diffInMillies,TimeUnit.MILLISECONDS);
}

然后你需要打电话来获取你的价值

getDateDiff(date1,date2,TimeUnit.MINUTES);

Answer 2

尝试这个

public double  findDifference(String start_date, String end_date)
{
    // SimpleDateFormat converts the
    // string format to date object
   SimpleDateFormat sdf = new SimpleDateFormat("dd-MM-yyyy HH:mm:ss");
   double diff = 0;
    // Try Block
    try {
        // parse method is used to parse
        // the text from a string to
        // produce the date
        Date d1 = sdf.parse(start_date);
        Date d2 = sdf.parse(end_date);
        // Calucalte time difference
        // in milliseconds
        long difference_In_Time = d2.getTime() - d1.getTime();

        // Calucalte time difference in
        // seconds, minutes, hours, years,
        // and days
        long difference_In_Seconds = (difference_In_Time / 1000) % 60;

        long difference_In_Minutes = (difference_In_Time / (1000 * 60)) % 60;

        long day   = (difference_In_Time / (1000 * 60 * 60)) % 24;

        long difference_In_Years = (difference_In_Time / (1000L * 60 * 60 * 24 * 365));

        long difference_In_Days = (difference_In_Time / (1000 * 60 * 60 * 24)) % 365;

        // Print the date difference in
        // years, in days, in hours, in
        // minutes, and in seconds
    }

    // Catch the Exception
    catch (ParseException e) {
        e.printStackTrace();
    }
    return diff;
}