48

我需要从毫秒到代表相同时间量的（小时，分钟，秒，毫秒）的元组。例如：

10799999ms = 2h 59m 59s 999ms

以下伪代码是我唯一能想到的：

# The division operator below returns the result as a rounded down integer
function to_tuple(x):
    h = x / (60*60*1000)
    x = x - h*(60*60*1000)
    m = x / (60*1000)
    x = x - m*(60*1000)
    s = x / 1000
    x = x - s*1000
    return (h,m,s,x)

我确信它必须可以做得更聪明/更优雅/更快/更紧凑。

4

10 回答 10

140

以下是我在 Java 中的做法：

int seconds = (int) (milliseconds / 1000) % 60 ;
int minutes = (int) ((milliseconds / (1000*60)) % 60);
int hours   = (int) ((milliseconds / (1000*60*60)) % 24);
于 2012-06-03T21:38:44.833 回答
7

好问题。是的，可以更有效地做到这一点。您的 CPU 可以在一次操作中提取两个整数之比的商和余数。在<stdlib.h>中，调用公开此 CPU 操作的函数div()。在你的伪代码中，你会像这样使用它：

function to_tuple(x):
    qr = div(x, 1000)
    ms = qr.rem
    qr = div(qr.quot, 60)
    s  = qr.rem
    qr = div(qr.quot, 60)
    m  = qr.rem
    h  = qr.quot

一个效率较低的答案将分别使用/and%运算符。但是，无论如何，如果您需要商和余数，那么您不妨调用更高效的div().

于 2012-06-03T21:39:50.897 回答
5

也许可以更短更优雅。但我做到了。

public String getHumanTimeFormatFromMilliseconds(String millisecondS){
    String message = "";
    long milliseconds = Long.valueOf(millisecondS);
    if (milliseconds >= 1000){
        int seconds = (int) (milliseconds / 1000) % 60;
        int minutes = (int) ((milliseconds / (1000 * 60)) % 60);
        int hours = (int) ((milliseconds / (1000 * 60 * 60)) % 24);
        int days = (int) (milliseconds / (1000 * 60 * 60 * 24));
        if((days == 0) && (hours != 0)){
            message = String.format("%d hours %d minutes %d seconds ago", hours, minutes, seconds);
        }else if((hours == 0) && (minutes != 0)){
            message = String.format("%d minutes %d seconds ago", minutes, seconds);
        }else if((days == 0) && (hours == 0) && (minutes == 0)){
            message = String.format("%d seconds ago", seconds);
        }else{
            message = String.format("%d days %d hours %d minutes %d seconds ago", days, hours, minutes, seconds);
        }
    } else{
        message = "Less than a second ago.";
    }
    return message;
}
于 2015-07-23T14:59:24.943 回答
3

不是很优雅，但会更短一些

function to_tuple(x):
   y = 60*60*1000
   h = x/y
   m = (x-(h*y))/(y/60)
   s = (x-(h*y)-(m*(y/60)))/1000
   mi = x-(h*y)-(m*(y/60))-(s*1000)

   return (h,m,s,mi)
于 2012-06-03T21:43:10.723 回答
0 
milliseconds = 12884983  // or x milliseconds
hr = 0
min = 0
sec = 0 
day = 0
while (milliseconds >= 1000) {
  milliseconds = (milliseconds - 1000)
  sec = sec + 1
  if (sec >= 60) min = min + 1
  if (sec == 60) sec = 0
  if (min >= 60) hr = hr + 1
  if (min == 60) min = 0
  if (hr >= 24) {
    hr = (hr - 24)
    day = day + 1
  }
}

我希望我的简短方法对您有所帮助

于 2016-12-04T14:26:42.453 回答
0

Kotlin 示例，其前导零的小时/分钟/秒小于 10。因此，如果您想在 UI 中将该值用作字符串，那么您最终会得到相同的持续时间。

这给出了 01:57:01 而不是 1:57:1，这可能会使 hh:mn:ss 注释感到困惑。

val timeInMilliSec = 45600030

val hours = timeInMilliSec.div(3600).rem(24)
val minutes = timeInMilliSec.div(60).rem(60)
val seconds = timeInMilliSec.rem(60)

val hoursFormatted = if (hours < 10) "0$hours" else "$hours"
val minutesFormatted = if (minutes < 10) "0$minutes" else "$minutes"
val secondsFormatted = if (seconds < 10) "0$seconds" else "$seconds"

"$hoursFormatted:$minutesFormatted:$secondsFormatted"
于 2021-08-02T12:17:28.483 回答
0 
milliseconds = x
total = 0
while (milliseconds >= 1000) {
  milliseconds = (milliseconds - 1000)
  total = total + 1
}
hr = 0
min = 0
while (total >= 60) {
  total = total - 60
  min = min + 1
  if (min >= 60) hr = hr + 1
  if (min == 60) min = 0
}
sec = total

这是常规的，但我认为这对你来说不是问题。方法工作完美。

于 2016-12-04T14:11:43.200 回答
0

基于 Valentinos 答案的 Arduino (c++) 版本

unsigned long timeNow = 0;
unsigned long mSecInHour = 3600000;
unsigned long TimeNow =0;
int millisecs =0;  
int seconds = 0;
byte minutes = 0;
byte hours = 0;

void setup() {
Serial.begin(9600);
Serial.println (""); // because arduino monitor gets confused with line 1
Serial.println ("hours:minutes:seconds.milliseconds:");
}

void loop() {
TimeNow = millis(); 
hours = TimeNow/mSecInHour;
minutes = (TimeNow-(hours*mSecInHour))/(mSecInHour/60);
seconds = (TimeNow-(hours*mSecInHour)-(minutes*(mSecInHour/60)))/1000;
millisecs = TimeNow-(hours*mSecInHour)-(minutes*(mSecInHour/60))-       (seconds*1000);

Serial.print(hours);  
Serial.print(":");
Serial.print(minutes);
Serial.print(":"); 
Serial.print(seconds); 
Serial.print("."); 
Serial.println(millisecs); 
}
于 2018-03-05T19:47:20.073 回答
0

只是另一个java示例：

long dayLength = 1000 * 60 * 60 * 24;
long dayMs = System.currentTimeMillis() % dayLength;
double percentOfDay = (double) dayMs / dayLength;
int hour = (int) (percentOfDay * 24);
int minute = (int) (percentOfDay * 24 * 60) % 60;
int second = (int) (percentOfDay * 24 * 60 * 60) % 60;

一个优点是你可以模拟更短的日子，如果你调整dayLength

于 2019-03-13T21:38:38.983 回答
0 
import java.text.SimpleDateFormat;
import java.util.Date;
import java.util.concurrent.TimeUnit;

public class MyTest {

    public static void main(String[] args) {
        long seconds = 360000;

        long days = TimeUnit.SECONDS.toDays(seconds);
        long hours = TimeUnit.SECONDS.toHours(seconds - TimeUnit.DAYS.toSeconds(days));

        System.out.println("days: " + days);
        System.out.println("hours: " + hours);
    }
}
于 2017-09-28T14:15:40.697 回答