我想使用System.currentTimeMillis()记录用户在程序中开始某项操作时的时间。当他完成时,我将从start变量中减去当前的System.currentTimeMillis(),并且我希望使用人类可读的格式显示他们所经过的时间,例如“XX小时,XX分钟,XX秒”,甚至“XX分钟,XX秒”,因为它不太可能花费某人一个小时。
最好的方法是什么?
我想使用System.currentTimeMillis()记录用户在程序中开始某项操作时的时间。当他完成时,我将从start变量中减去当前的System.currentTimeMillis(),并且我希望使用人类可读的格式显示他们所经过的时间,例如“XX小时,XX分钟,XX秒”,甚至“XX分钟,XX秒”,因为它不太可能花费某人一个小时。
最好的方法是什么?
当前回答
DurationFormatUtils.formatDurationHMS(长)
其他回答
基于@siddhadev的回答,我写了一个将毫秒转换为格式化字符串的函数:
/**
* Convert a millisecond duration to a string format
*
* @param millis A duration to convert to a string form
* @return A string of the form "X Days Y Hours Z Minutes A Seconds".
*/
public static String getDurationBreakdown(long millis) {
if(millis < 0) {
throw new IllegalArgumentException("Duration must be greater than zero!");
}
long days = TimeUnit.MILLISECONDS.toDays(millis);
millis -= TimeUnit.DAYS.toMillis(days);
long hours = TimeUnit.MILLISECONDS.toHours(millis);
millis -= TimeUnit.HOURS.toMillis(hours);
long minutes = TimeUnit.MILLISECONDS.toMinutes(millis);
millis -= TimeUnit.MINUTES.toMillis(minutes);
long seconds = TimeUnit.MILLISECONDS.toSeconds(millis);
StringBuilder sb = new StringBuilder(64);
sb.append(days);
sb.append(" Days ");
sb.append(hours);
sb.append(" Hours ");
sb.append(minutes);
sb.append(" Minutes ");
sb.append(seconds);
sb.append(" Seconds");
return(sb.toString());
}
如果你知道时间差将小于一个小时,那么你可以使用以下代码:
Calendar c1 = Calendar.getInstance();
Calendar c2 = Calendar.getInstance();
c2.add(Calendar.MINUTE, 51);
long diff = c2.getTimeInMillis() - c1.getTimeInMillis();
c2.set(Calendar.MINUTE, 0);
c2.set(Calendar.HOUR, 0);
c2.set(Calendar.SECOND, 0);
DateFormat df = new SimpleDateFormat("mm:ss");
long diff1 = c2.getTimeInMillis() + diff;
System.out.println(df.format(new Date(diff1)));
结果是:51:00
使用java . util . concurrent。TimeUnit,使用这个简单的方法:
private static long timeDiff(Date date, Date date2, TimeUnit unit) {
long milliDiff=date2.getTime()-date.getTime();
long unitDiff = unit.convert(milliDiff, TimeUnit.MILLISECONDS);
return unitDiff;
}
例如:
SimpleDateFormat sdf = new SimpleDateFormat("yy/MM/dd HH:mm:ss");
Date firstDate = sdf.parse("06/24/2017 04:30:00");
Date secondDate = sdf.parse("07/24/2017 05:00:15");
Date thirdDate = sdf.parse("06/24/2017 06:00:15");
System.out.println("days difference: "+timeDiff(firstDate,secondDate,TimeUnit.DAYS));
System.out.println("hours difference: "+timeDiff(firstDate,thirdDate,TimeUnit.HOURS));
System.out.println("minutes difference: "+timeDiff(firstDate,thirdDate,TimeUnit.MINUTES));
System.out.println("seconds difference: "+timeDiff(firstDate,thirdDate,TimeUnit.SECONDS));
下面是一个基于Brent Nash的答案,希望有帮助!
public static String getDurationBreakdown(long millis)
{
String[] units = {" Days ", " Hours ", " Minutes ", " Seconds "};
Long[] values = new Long[units.length];
if(millis < 0)
{
throw new IllegalArgumentException("Duration must be greater than zero!");
}
values[0] = TimeUnit.MILLISECONDS.toDays(millis);
millis -= TimeUnit.DAYS.toMillis(values[0]);
values[1] = TimeUnit.MILLISECONDS.toHours(millis);
millis -= TimeUnit.HOURS.toMillis(values[1]);
values[2] = TimeUnit.MILLISECONDS.toMinutes(millis);
millis -= TimeUnit.MINUTES.toMillis(values[2]);
values[3] = TimeUnit.MILLISECONDS.toSeconds(millis);
StringBuilder sb = new StringBuilder(64);
boolean startPrinting = false;
for(int i = 0; i < units.length; i++){
if( !startPrinting && values[i] != 0)
startPrinting = true;
if(startPrinting){
sb.append(values[i]);
sb.append(units[i]);
}
}
return(sb.toString());
}
使用java.util.concurrent.TimeUnit类:
String.format("%d min, %d sec",
TimeUnit.MILLISECONDS.toMinutes(millis),
TimeUnit.MILLISECONDS.toSeconds(millis) -
TimeUnit.MINUTES.toSeconds(TimeUnit.MILLISECONDS.toMinutes(millis))
);
注意:TimeUnit是Java 1.5规范的一部分,但是toMinutes是在Java 1.6中添加的。
为0-9的值添加前导零,只需执行以下操作:
String.format("%02d min, %02d sec",
TimeUnit.MILLISECONDS.toMinutes(millis),
TimeUnit.MILLISECONDS.toSeconds(millis) -
TimeUnit.MINUTES.toSeconds(TimeUnit.MILLISECONDS.toMinutes(millis))
);
如果TimeUnit或toMinutes不受支持(例如在API版本9之前的Android上),使用以下公式:
int seconds = (int) (milliseconds / 1000) % 60 ;
int minutes = (int) ((milliseconds / (1000*60)) % 60);
int hours = (int) ((milliseconds / (1000*60*60)) % 24);
//etc...