我想要这样的东西:

public class Stream
{
    public startTime;
    public endTime;

    public getDuration()
    {
        return startTime - endTime;
    }
}

同样重要的是,例如,如果startTime是23:00,endTime是1:00,则持续时间为2:00。

为了在Java中实现这一点,应该使用哪些类型?


当前回答

不幸的是,目前发布的十个答案中没有一个是完全正确的。

如果您正在测量经过的时间,并且希望它是正确的,那么必须使用System.nanoTime()。您不能使用System.currentTimeMillis(),除非您不介意结果是错误的。

The purpose of nanoTime is to measure elapsed time, and the purpose of currentTimeMillis is to measure wall-clock time. You can't use the one for the other purpose. The reason is that no computer's clock is perfect; it always drifts and occasionally needs to be corrected. This correction might either happen manually, or in the case of most machines, there's a process that runs and continually issues small corrections to the system clock ("wall clock"). These tend to happen often. Another such correction happens whenever there is a leap second.

由于nanoTime的目的是测量经过的时间,因此它不受任何这些小修正的影响。它是你想要使用的。当前使用currentTimeMillis进行的任何计时都将关闭——甚至可能为负值。

您可能会说,“这听起来似乎并没有那么重要”,对此我说,也许不是,但总的来说,正确的代码不就是比错误的代码好吗?此外,nanoTime打字时间更短。

之前发布的关于nanoTime通常只有微秒精度的免责声明是有效的。此外,它可能需要超过一微秒的时间来调用,这取决于环境(另一个也是如此),所以不要期望对非常非常小的间隔进行正确的计时。

其他回答

我建立了一个格式化函数,基于我偷来的东西。我需要一种方法来“分析”日志消息中的内容,所以我需要一个固定长度的持续时间消息。

public static String GetElapsed(long aInitialTime, long aEndTime, boolean aIncludeMillis)
{
  StringBuffer elapsed = new StringBuffer();

  Map<String, Long> units = new HashMap<String, Long>();

  long milliseconds = aEndTime - aInitialTime;

  long seconds = milliseconds / 1000;
  long minutes = milliseconds / (60 * 1000);
  long hours = milliseconds / (60 * 60 * 1000);
  long days = milliseconds / (24 * 60 * 60 * 1000);

  units.put("milliseconds", milliseconds);
  units.put("seconds", seconds);
  units.put("minutes", minutes);
  units.put("hours", hours);
  units.put("days", days);

  if (days > 0)
  {
    long leftoverHours = hours % 24;
    units.put("hours", leftoverHours);
  }

  if (hours > 0)
  {
    long leftoeverMinutes = minutes % 60;
    units.put("minutes", leftoeverMinutes);
  }

  if (minutes > 0)
  {
    long leftoverSeconds = seconds % 60;
    units.put("seconds", leftoverSeconds);
  }

  if (seconds > 0)
  {
    long leftoverMilliseconds = milliseconds % 1000;
    units.put("milliseconds", leftoverMilliseconds);
  }

  elapsed.append(PrependZeroIfNeeded(units.get("days")) + " days ")
      .append(PrependZeroIfNeeded(units.get("hours")) + " hours ")
      .append(PrependZeroIfNeeded(units.get("minutes")) + " minutes ")
      .append(PrependZeroIfNeeded(units.get("seconds")) + " seconds ")
      .append(PrependZeroIfNeeded(units.get("milliseconds")) + " ms");

  return elapsed.toString();

}

private static String PrependZeroIfNeeded(long aValue)
{
  return aValue < 10 ? "0" + aValue : Long.toString(aValue);
}

和一个测试类:

import java.util.Calendar;
import java.util.Date;
import java.util.GregorianCalendar;

import junit.framework.TestCase;

public class TimeUtilsTest extends TestCase
{

  public void testGetElapsed()
  {
    long start = System.currentTimeMillis();
    GregorianCalendar calendar = (GregorianCalendar) Calendar.getInstance();
    calendar.setTime(new Date(start));

    calendar.add(Calendar.MILLISECOND, 610);
    calendar.add(Calendar.SECOND, 35);
    calendar.add(Calendar.MINUTE, 5);
    calendar.add(Calendar.DAY_OF_YEAR, 5);

    long end = calendar.getTimeInMillis();

    assertEquals("05 days 00 hours 05 minutes 35 seconds 610 ms", TimeUtils.GetElapsed(start, end, true));

  }

}

如果目的只是将粗略的计时信息打印到程序日志中,那么Java项目的简单解决方案不是编写自己的秒表或计时器类,而是使用Apache Commons Lang中的org.apache.commons.lang.time.StopWatch类。

final StopWatch stopwatch = new StopWatch();
stopwatch.start();
LOGGER.debug("Starting long calculations: {}", stopwatch);
...
LOGGER.debug("Time after key part of calcuation: {}", stopwatch);
...
LOGGER.debug("Finished calculating {}", stopwatch);

你的新类:

public class TimeWatch {    
    long starts;

    public static TimeWatch start() {
        return new TimeWatch();
    }

    private TimeWatch() {
        reset();
    }

    public TimeWatch reset() {
        starts = System.currentTimeMillis();
        return this;
    }

    public long time() {
        long ends = System.currentTimeMillis();
        return ends - starts;
    }

    public long time(TimeUnit unit) {
        return unit.convert(time(), TimeUnit.MILLISECONDS);
    }
}

用法:

    TimeWatch watch = TimeWatch.start();
    // do something
    long passedTimeInMs = watch.time();
    long passedTimeInSeconds = watch.time(TimeUnit.SECONDS);

之后,所经过的时间可以转换为任何你喜欢的格式,例如日历

Greetz, 查德

我发现这段代码在计时时很有用:

public class Et {
    public Et() {
    reset();
    }
    public void reset() {
    t0=System.nanoTime();
    }
    public long t0() {
        return t0;
    }
    public long dt() {
        return System.nanoTime()-t0();
    }
    public double etms() {
    return etms(dt());
    }
    @Override public String toString() {
        return etms()+" ms.";
    }
    public static double etms(long dt) {
        return dt/1000000.; // 1_000_000. breaks cobertura
    }
    private Long t0;
}

不幸的是,目前发布的十个答案中没有一个是完全正确的。

如果您正在测量经过的时间,并且希望它是正确的,那么必须使用System.nanoTime()。您不能使用System.currentTimeMillis(),除非您不介意结果是错误的。

The purpose of nanoTime is to measure elapsed time, and the purpose of currentTimeMillis is to measure wall-clock time. You can't use the one for the other purpose. The reason is that no computer's clock is perfect; it always drifts and occasionally needs to be corrected. This correction might either happen manually, or in the case of most machines, there's a process that runs and continually issues small corrections to the system clock ("wall clock"). These tend to happen often. Another such correction happens whenever there is a leap second.

由于nanoTime的目的是测量经过的时间,因此它不受任何这些小修正的影响。它是你想要使用的。当前使用currentTimeMillis进行的任何计时都将关闭——甚至可能为负值。

您可能会说,“这听起来似乎并没有那么重要”,对此我说,也许不是,但总的来说,正确的代码不就是比错误的代码好吗?此外,nanoTime打字时间更短。

之前发布的关于nanoTime通常只有微秒精度的免责声明是有效的。此外,它可能需要超过一微秒的时间来调用,这取决于环境(另一个也是如此),所以不要期望对非常非常小的间隔进行正确的计时。