两个日期之间的java差异

Question

我需要帮助。 我有两个约会。

private Date endDate;
private Date now;

public String getRemainingTime(endDate) {
   now = new Date();

   //logic

}

还有我有一种方法应该在String formate中返回两个日期之间的差异，如果差异超过1天-例如“ 1天15分钟”；

如果时差大于1小时但小于1天-例如“ 1小时13分钟”；

如果时差少于1小时-“ 35分39秒”，则类似以下内容...

请帮助我，我对java不熟悉。

Answer 1

如果建议的评论链接不起作用，请尝试以下方法：

Date endDate;
         Date now;
         StringBuilder sb = new StringBuilder();
        //timestamp difference in millis 
        Long diff = endaDate.getTime() - now.getTime();
        //get the seconds
        long seconds = diff / 1000;
        //get the minutes
        long minutes = diff / (1000 * 60);
        //get the hours
        long hrs = diff / (1000 * 60 * 60);
        if (hrs > 0) {
            sb.append(hrs + " hrs");
            if (minutes % 60 > 0) {
                sb.append(", " + minutes % 60 + " mins");
            }
            if (seconds % 60 > 0) {
                sb.append(", " + seconds % 60 + " secs");
            }
        } else if (minutes % 60 > 0) {
            sb.append(minutes % 60 + " mins");
            if (seconds % 60 > 0) {
                sb.append(", " + seconds % 60 + " secs");
            }
        } else if (seconds % 60 > 0) {
            sb.append(seconds % 60 + " secs");
        } else {
            sb.append("00");
        }
        System.out.println(sb.toString());

`

Answer 2

更新
1）该代码将完成这项工作（编译器将包装常量）：

public String getIntervalAsString(Date date1, Date date2) {
        String format;

        long dT = date2.getTime() - date1.getTime();
        if (dT < 1000 * 60)
            format = "s 'sec'";
        else if (dT < 1000 * 60 * 60)
            format = "m 'min' s 'sec'";
        else if (dT  < 1000 * 60 * 60 * 24)
            format = "h 'hour(s)' m 'min'";
        else if (dT < 1000 * 60 * 60 * 24 * 365)
            format = "d 'day(s)' h 'hour(s)'";
        else
            format = "'more than a year'";

        SimpleDateFormat formatter = new SimpleDateFormat(format);

        return formatter.format(new Date(dT));
    }

2）您可以在这里尝试不同的模式

Answer 3

private static final long MILLIS_PER_SECOND = 1000L;
private static final long MILLIS_PER_MINUTE = MILLIS_PER_SECOND * 60L;
private static final long MILLIS_PER_HOUR = MILLIS_PER_MINUTE * 60L;
private static final long MILLIS_PER_DAY = MILLIS_PER_HOUR * 24L;

public String getRemainingTime(Date endDate) {
    long remain = endDate.getTime() - System.currentTimeMillis();
    if (remain <= 0L) {
        return "pass";
    }
    long days = remain / MILLIS_PER_DAY;
    long hours = (remain % MILLIS_PER_DAY) / MILLIS_PER_HOUR;
    long minutes = (remain % MILLIS_PER_HOUR) / MILLIS_PER_MINUTE;
    long seconds = (remain % MILLIS_PER_MINUTE) / MILLIS_PER_SECOND;
    StringBuilder sb = new StringBuilder();
    if (days > 0L) {
        sb.append(days).append(" day ");
    }
    if (hours > 0L || days > 0L) {
        sb.append(hours).append(" hour ");
    }
    if (minutes > 0L || days > 0L || hours > 0L) {
        sb.append(minutes).append(" min ");
    }
    if (seconds > 0L) {
        sb.append(seconds).append(" sec");
    }
    return sb.toString();
}

Answer 4

Joda-Time库可能符合您的需求：

DateTime now
DateTime endDate;
Period p = new Period(now, endDate);
long hours = p.getHours();
long minutes = p.getMinutes();