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?
嗯……一秒是多少毫秒?一分钟后呢?除法没那么难。
int seconds = (int) ((milliseconds / 1000) % 60);
int minutes = (int) ((milliseconds / 1000) / 60);
这样持续几小时、几天、几周、几个月、一年、几十年,等等。
手工划分,或者使用SimpleDateFormat API。
long start = System.currentTimeMillis();
// do your work...
long elapsed = System.currentTimeMillis() - start;
DateFormat df = new SimpleDateFormat("HH 'hours', mm 'mins,' ss 'seconds'");
df.setTimeZone(TimeZone.getTimeZone("GMT+0"));
System.out.println(df.format(new Date(elapsed)));
Edit by Bombe:评论中显示,这种方法只适用于较短的持续时间(即少于一天)。
我不会为此引入额外的依赖项(毕竟,除法并不那么难),但如果您无论如何都在使用Commons Lang,还有DurationFormatUtils。
用法示例(从这里改编):
import org.apache.commons.lang3.time.DurationFormatUtils
public String getAge(long value) {
long currentTime = System.currentTimeMillis();
long age = currentTime - value;
String ageString = DurationFormatUtils.formatDuration(age, "d") + "d";
if ("0d".equals(ageString)) {
ageString = DurationFormatUtils.formatDuration(age, "H") + "h";
if ("0h".equals(ageString)) {
ageString = DurationFormatUtils.formatDuration(age, "m") + "m";
if ("0m".equals(ageString)) {
ageString = DurationFormatUtils.formatDuration(age, "s") + "s";
if ("0s".equals(ageString)) {
ageString = age + "ms";
}
}
}
}
return ageString;
}
例子:
long lastTime = System.currentTimeMillis() - 2000;
System.out.println("Elapsed time: " + getAge(lastTime));
//Output: 2s
注意:从两个LocalDateTime对象中获取millis可以使用:
long age = ChronoUnit.MILLIS.between(initTime, LocalDateTime.now())
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...
我认为最好的办法是:
String.format("%d min, %d sec",
TimeUnit.MILLISECONDS.toSeconds(length)/60,
TimeUnit.MILLISECONDS.toSeconds(length) % 60 );
对于一小时以内的小时间,我更喜欢:
long millis = ...
System.out.printf("%1$TM:%1$TS", millis);
// or
String str = String.format("%1$TM:%1$TS", millis);
对于较长的间隔:
private static final long HOUR = TimeUnit.HOURS.toMillis(1);
...
if (millis < HOUR) {
System.out.printf("%1$TM:%1$TS%n", millis);
} else {
System.out.printf("%d:%2$TM:%2$TS%n", millis / HOUR, millis % HOUR);
}
只是为了补充更多信息 如果你想格式化为:HH:mm:ss
0 <= HH <= infinite
0 <= mm < 60
0 <= ss < 60
用这个:
int h = (int) ((startTimeInMillis / 1000) / 3600);
int m = (int) (((startTimeInMillis / 1000) / 60) % 60);
int s = (int) ((startTimeInMillis / 1000) % 60);
我只是现在有这个问题,并解决了这个问题
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());
}
对于正确的字符串(“1小时,3秒”,“3分钟”,而不是“0小时,0分钟,3秒”),我写这样的代码:
int seconds = (int)(millis / 1000) % 60 ;
int minutes = (int)((millis / (1000*60)) % 60);
int hours = (int)((millis / (1000*60*60)) % 24);
int days = (int)((millis / (1000*60*60*24)) % 365);
int years = (int)(millis / 1000*60*60*24*365);
ArrayList<String> timeArray = new ArrayList<String>();
if(years > 0)
timeArray.add(String.valueOf(years) + "y");
if(days > 0)
timeArray.add(String.valueOf(days) + "d");
if(hours>0)
timeArray.add(String.valueOf(hours) + "h");
if(minutes>0)
timeArray.add(String.valueOf(minutes) + "min");
if(seconds>0)
timeArray.add(String.valueOf(seconds) + "sec");
String time = "";
for (int i = 0; i < timeArray.size(); i++)
{
time = time + timeArray.get(i);
if (i != timeArray.size() - 1)
time = time + ", ";
}
if (time == "")
time = "0 sec";
long time = 1536259;
return (new SimpleDateFormat("mm:ss:SSS")).format(new Date(time));
打印:
25:36:259
long startTime = System.currentTimeMillis();
// do your work...
long endTime=System.currentTimeMillis();
long diff=endTime-startTime;
long hours=TimeUnit.MILLISECONDS.toHours(diff);
diff=diff-(hours*60*60*1000);
long min=TimeUnit.MILLISECONDS.toMinutes(diff);
diff=diff-(min*60*1000);
long seconds=TimeUnit.MILLISECONDS.toSeconds(diff);
//hour, min and seconds variables contains the time elapsed on your work
适用于API 9以下的Android
(String.format("%d hr %d min, %d sec", millis/(1000*60*60), (millis%(1000*60*60))/(1000*60), ((millis%(1000*60*60))%(1000*60))/1000))
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());
}
乔达时间
使用Joda-Time:
DateTime startTime = new DateTime();
// do something
DateTime endTime = new DateTime();
Duration duration = new Duration(startTime, endTime);
Period period = duration.toPeriod().normalizedStandard(PeriodType.time());
System.out.println(PeriodFormat.getDefault().print(period));
使用java。Java 8中的时间包:
Instant start = Instant.now();
Thread.sleep(63553);
Instant end = Instant.now();
System.out.println(Duration.between(start, end));
输出为ISO 8601 Duration格式:pt1m3.53s(1分3.553秒)。
最短的解决方案:
这可能是最短的一个,也涉及到时区。
System.out.printf("%tT", millis-TimeZone.getDefault().getRawOffset());
输出例如:
00:18:32
解释:
%tT是24小时时钟格式为%tH:%tM:%tS的时间。
%tT也接受长时间作为输入,因此不需要创建Date。Printf()将简单地以毫秒为单位打印指定的时间,但在当前时区中,因此我们必须减去当前时区的原始偏移量,以便0毫秒将是0小时,而不是当前时区的时间偏移值。
注意#1:如果你需要一个字符串形式的结果,你可以这样得到它:
String t = String.format("%tT", millis-TimeZone.getDefault().getRawOffset());
注意#2:只有当millis小于一天时,才会给出正确的结果,因为一天部分不包括在输出中。
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
这个答案与上面的一些答案相似。然而,我觉得这将是有益的,因为与其他答案不同,这将删除任何额外的逗号或空格,并处理缩写。
/**
* Converts milliseconds to "x days, x hours, x mins, x secs"
*
* @param millis
* The milliseconds
* @param longFormat
* {@code true} to use "seconds" and "minutes" instead of "secs" and "mins"
* @return A string representing how long in days/hours/minutes/seconds millis is.
*/
public static String millisToString(long millis, boolean longFormat) {
if (millis < 1000) {
return String.format("0 %s", longFormat ? "seconds" : "secs");
}
String[] units = {
"day", "hour", longFormat ? "minute" : "min", longFormat ? "second" : "sec"
};
long[] times = new long[4];
times[0] = TimeUnit.DAYS.convert(millis, TimeUnit.MILLISECONDS);
millis -= TimeUnit.MILLISECONDS.convert(times[0], TimeUnit.DAYS);
times[1] = TimeUnit.HOURS.convert(millis, TimeUnit.MILLISECONDS);
millis -= TimeUnit.MILLISECONDS.convert(times[1], TimeUnit.HOURS);
times[2] = TimeUnit.MINUTES.convert(millis, TimeUnit.MILLISECONDS);
millis -= TimeUnit.MILLISECONDS.convert(times[2], TimeUnit.MINUTES);
times[3] = TimeUnit.SECONDS.convert(millis, TimeUnit.MILLISECONDS);
StringBuilder s = new StringBuilder();
for (int i = 0; i < 4; i++) {
if (times[i] > 0) {
s.append(String.format("%d %s%s, ", times[i], units[i], times[i] == 1 ? "" : "s"));
}
}
return s.toString().substring(0, s.length() - 2);
}
/**
* Converts milliseconds to "x days, x hours, x mins, x secs"
*
* @param millis
* The milliseconds
* @return A string representing how long in days/hours/mins/secs millis is.
*/
public static String millisToString(long millis) {
return millisToString(millis, false);
}
我修改了@MyKuLLSKI的回答,增加了多元化支持。我去掉了几秒,因为我不需要它们,如果你需要的话,可以重新添加。
public static String intervalToHumanReadableTime(int intervalMins) {
if(intervalMins <= 0) {
return "0";
} else {
long intervalMs = intervalMins * 60 * 1000;
long days = TimeUnit.MILLISECONDS.toDays(intervalMs);
intervalMs -= TimeUnit.DAYS.toMillis(days);
long hours = TimeUnit.MILLISECONDS.toHours(intervalMs);
intervalMs -= TimeUnit.HOURS.toMillis(hours);
long minutes = TimeUnit.MILLISECONDS.toMinutes(intervalMs);
StringBuilder sb = new StringBuilder(12);
if (days >= 1) {
sb.append(days).append(" day").append(pluralize(days)).append(", ");
}
if (hours >= 1) {
sb.append(hours).append(" hour").append(pluralize(hours)).append(", ");
}
if (minutes >= 1) {
sb.append(minutes).append(" minute").append(pluralize(minutes));
} else {
sb.delete(sb.length()-2, sb.length()-1);
}
return(sb.toString());
}
}
public static String pluralize(long val) {
return (Math.round(val) > 1 ? "s" : "");
}
我在另一个答案中提到了这一点,但你可以这样做:
public static Map<TimeUnit,Long> computeDiff(Date date1, Date date2) {
long diffInMillies = date2.getTime() - date1.getTime();
List<TimeUnit> units = new ArrayList<TimeUnit>(EnumSet.allOf(TimeUnit.class));
Collections.reverse(units);
Map<TimeUnit,Long> result = new LinkedHashMap<TimeUnit,Long>();
long milliesRest = diffInMillies;
for ( TimeUnit unit : units ) {
long diff = unit.convert(milliesRest,TimeUnit.MILLISECONDS);
long diffInMilliesForUnit = unit.toMillis(diff);
milliesRest = milliesRest - diffInMilliesForUnit;
result.put(unit,diff);
}
return result;
}
输出类似Map:{DAYS=1, HOURS=3, MINUTES=46, SECONDS=40, MILLISECONDS=0, MICROSECONDS=0, NANOSECONDS=0},单位是有序的。
由您决定如何根据目标语言环境对这些数据进行国际化。
回顾@brent-nash的贡献,我们可以使用模函数代替减法并使用String。结果字符串的格式方法:
/**
* 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 B Milliseconds".
*/
public static String getDurationBreakdown(long millis) {
if (millis < 0) {
throw new IllegalArgumentException("Duration must be greater than zero!");
}
long days = TimeUnit.MILLISECONDS.toDays(millis);
long hours = TimeUnit.MILLISECONDS.toHours(millis) % 24;
long minutes = TimeUnit.MILLISECONDS.toMinutes(millis) % 60;
long seconds = TimeUnit.MILLISECONDS.toSeconds(millis) % 60;
long milliseconds = millis % 1000;
return String.format("%d Days %d Hours %d Minutes %d Seconds %d Milliseconds",
days, hours, minutes, seconds, milliseconds);
}
我的简单计算是:
String millisecToTime(int millisec) {
int sec = millisec/1000;
int second = sec % 60;
int minute = sec / 60;
if (minute >= 60) {
int hour = minute / 60;
minute %= 60;
return hour + ":" + (minute < 10 ? "0" + minute : minute) + ":" + (second < 10 ? "0" + second : second);
}
return minute + ":" + (second < 10 ? "0" + second : second);
}
快乐编码:)
这在Java 9中更容易:
Duration elapsedTime = Duration.ofMillis(millisDiff );
String humanReadableElapsedTime = String.format(
"%d hours, %d mins, %d seconds",
elapsedTime.toHours(),
elapsedTime.toMinutesPart(),
elapsedTime.toSecondsPart());
这将生成像0小时39分9秒这样的字符串。
如果你想在格式化前舍入整秒:
elapsedTime = elapsedTime.plusMillis(500).truncatedTo(ChronoUnit.SECONDS);
如果小时数为0,则省略:
long hours = elapsedTime.toHours();
String humanReadableElapsedTime;
if (hours == 0) {
humanReadableElapsedTime = String.format(
"%d mins, %d seconds",
elapsedTime.toMinutesPart(),
elapsedTime.toSecondsPart());
} else {
humanReadableElapsedTime = String.format(
"%d hours, %d mins, %d seconds",
hours,
elapsedTime.toMinutesPart(),
elapsedTime.toSecondsPart());
}
现在我们有39分9秒。
要打印前导0的分钟和秒,使它们始终是两个数字,只需在相关的格式说明符中插入02,如下所示:
String humanReadableElapsedTime = String.format(
"%d hours, %02d mins, %02d seconds",
elapsedTime.toHours(),
elapsedTime.toMinutesPart(),
elapsedTime.toSecondsPart());
现在我们有0小时39分09秒。
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));
有个问题。当毫秒为59999时,实际上是1分钟,但它将被计算为59秒,损失了999毫秒。
以下是根据之前的答案修改后的版本,可以解决这个损失:
public static String formatTime(long millis) {
long seconds = Math.round((double) millis / 1000);
long hours = TimeUnit.SECONDS.toHours(seconds);
if (hours > 0)
seconds -= TimeUnit.HOURS.toSeconds(hours);
long minutes = seconds > 0 ? TimeUnit.SECONDS.toMinutes(seconds) : 0;
if (minutes > 0)
seconds -= TimeUnit.MINUTES.toSeconds(minutes);
return hours > 0 ? String.format("%02d:%02d:%02d", hours, minutes, seconds) : String.format("%02d:%02d", minutes, seconds);
}
首先,System.currentTimeMillis()和Instant.now()对于计时来说并不理想。它们都报告了挂钟时间,而计算机并不能准确地知道这个时间,而且它可能会不规律地移动,例如,如果NTP守护进程更正了系统时间,就会往回走。如果计时发生在一台机器上,那么应该使用System.nanoTime()。
其次,从Java 8开始,Java .time. duration是表示持续时间的最佳方式:
long start = System.nanoTime();
// do things...
long end = System.nanoTime();
Duration duration = Duration.ofNanos(end - start);
System.out.println(duration); // Prints "PT18M19.511627776S"
System.out.printf("%d Hours %d Minutes %d Seconds%n",
duration.toHours(), duration.toMinutes() % 60, duration.getSeconds() % 60);
// prints "0 Hours 18 Minutes 19 Seconds"
对于那些寻找Kotlin代码的人:
fun converter(millis: Long): String =
String.format(
"%02d : %02d : %02d",
TimeUnit.MILLISECONDS.toHours(millis),
TimeUnit.MILLISECONDS.toMinutes(millis) - TimeUnit.HOURS.toMinutes(
TimeUnit.MILLISECONDS.toHours(millis)
),
TimeUnit.MILLISECONDS.toSeconds(millis) - TimeUnit.MINUTES.toSeconds(
TimeUnit.MILLISECONDS.toMinutes(millis)
)
)
示例输出:09:10:26
这个主题已经被很好地覆盖了,我只是想分享我的函数,也许你可以使用这些函数,而不是导入整个库。
public long getSeconds(ms) {
return (ms/1000%60);
}
public long getMinutes(ms) {
return (ms/(1000*60)%60);
}
public long getHours(ms) {
return ((ms/(1000*60*60))%24);
}