如何将毫秒转换为人类可读的形式?


118

我需要将任意毫秒数转换为天,小时,分钟秒。

例如:10天5小时13分钟1秒。


“我使用的语言没有内置此语言,否则我会使用它。” 我觉得很难理解。什么语言?什么操作系统?
S.Lott

ActionScript,任何操作系统,都具有糟糕的日期/时间支持
FlySwat

3
我不知道有什么语言能满足他的要求,也看不出有什么理由。一些非常简单的除法/模数数学就可以很好地得到答案。
基普(Kip)

1
并非所有年份都有相同的天数,因此您必须说明那是哪个时期。或者,也许您只想在“标准”年份(365年左右)购买?
米兰·巴布斯科夫,

@Kip:明白了-误解了问题-想到的是以毫秒为单位的操作系统时间戳。没有增量时间或间隔。诱惑编辑的问题...
美国洛特

Answers:


226

好吧,既然没有其他人加紧努力,我将编写简单的代码来做到这一点:

x = ms / 1000
seconds = x % 60
x /= 60
minutes = x % 60
x /= 60
hours = x % 24
x /= 24
days = x

我很高兴你停了几天,没问了几个月。:)

注意,在上面,假定/代表截断整数除法。如果使用/表示浮点除法的语言使用此代码,则需要根据需要手动截断除法的结果。


2
刚刚在闪光灯功能中使用过。谢谢!(为简化起见,已投票)
Makram Saleh,2009年

2
它不能正常工作。使用除数时应使用parseInt,否则将看到长浮点值。请参阅下面的答案,以获得更全面的解决方案。
拉吉夫2012年

17
@Greg Hewgill 我很高兴您在几天内停下来,而且几个月都没问。:)哈哈:)
moshfiqur 2012年

58

令A为毫秒数。然后您有:

seconds=(A/1000)%60
minutes=(A/(1000*60))%60
hours=(A/(1000*60*60))%24

以此类推(%是模运算符)。

希望这可以帮助。


@sabbibJAVA 24应该已经工作了。你用什么语言?如果进行/浮点除法,则需要截断该值。在其他答案中假定/正在执行整数除法。
Brian J

24

以下两个解决方案都使用javascript(我不知道该解决方案与语言无关!)。如果捕获持续时间,则两种解决方案都需要扩展> 1 month

解决方案1:使用日期对象

var date = new Date(536643021);
var str = '';
str += date.getUTCDate()-1 + " days, ";
str += date.getUTCHours() + " hours, ";
str += date.getUTCMinutes() + " minutes, ";
str += date.getUTCSeconds() + " seconds, ";
str += date.getUTCMilliseconds() + " millis";
console.log(str);

给出:

"6 days, 5 hours, 4 minutes, 3 seconds, 21 millis"

库是有帮助的,但是当您可以重新发明轮子时,为什么还要使用库呢!:)

解决方案2:编写自己的解析器

var getDuration = function(millis){
    var dur = {};
    var units = [
        {label:"millis",    mod:1000},
        {label:"seconds",   mod:60},
        {label:"minutes",   mod:60},
        {label:"hours",     mod:24},
        {label:"days",      mod:31}
    ];
    // calculate the individual unit values...
    units.forEach(function(u){
        millis = (millis - (dur[u.label] = (millis % u.mod))) / u.mod;
    });
    // convert object to a string representation...
    var nonZero = function(u){ return dur[u.label]; };
    dur.toString = function(){
        return units
            .reverse()
            .filter(nonZero)
            .map(function(u){
                return dur[u.label] + " " + (dur[u.label]==1?u.label.slice(0,-1):u.label);
            })
            .join(', ');
    };
    return dur;
};

创建一个“ duration”对象,其中包含您需要的任何字段。格式化时间戳变得非常简单...

console.log(getDuration(536643021).toString());

给出:

"6 days, 5 hours, 4 minutes, 3 seconds, 21 millis"

更改该行以获取单数和复数 return dur[u.label] + " " + (dur[u.label]==1?u.label.slice(0,-1):u.label);
Phillip Kamikaze

1
@PhillipKamikaze感谢菲利普!我已经采纳了你的建议。
尼克·格莱利

您可能不希望显示具有零值的线段,因此可以添加以下过滤器... var nonZero = function(u){ return !u.startsWith("0"); }; // convert object to a string representation... dur.toString = function(){ return units.reverse().map(function(u){ return dur[u.label] + " " + (dur[u.label]==1?u.label.slice(0,-1):u.label); }).filter(nonZero).join(', '); };
Ruslan Ulanov

1
谢谢@RuslanUlanov!我已将其添加到示例中(尽管稍作修改以检查数字是否为“真”)。
尼克·格莱利


7

您应该使用所使用的任何语言的datetime函数,但是,为了好玩,这里是代码:

int milliseconds = someNumber;

int seconds = milliseconds / 1000;

int minutes = seconds / 60;

seconds %= 60;

int hours = minutes / 60;

minutes %= 60;

int days = hours / 24;

hours %= 24;

4

这是我写的一种方法。花费integer milliseconds value并返回human-readable String

public String convertMS(int ms) {
    int seconds = (int) ((ms / 1000) % 60);
    int minutes = (int) (((ms / 1000) / 60) % 60);
    int hours = (int) ((((ms / 1000) / 60) / 60) % 24);

    String sec, min, hrs;
    if(seconds<10)  sec="0"+seconds;
    else            sec= ""+seconds;
    if(minutes<10)  min="0"+minutes;
    else            min= ""+minutes;
    if(hours<10)    hrs="0"+hours;
    else            hrs= ""+hours;

    if(hours == 0)  return min+":"+sec;
    else    return hrs+":"+min+":"+sec;

}

4
function convertTime(time) {        
    var millis= time % 1000;
    time = parseInt(time/1000);
    var seconds = time % 60;
    time = parseInt(time/60);
    var minutes = time % 60;
    time = parseInt(time/60);
    var hours = time % 24;
    var out = "";
    if(hours && hours > 0) out += hours + " " + ((hours == 1)?"hr":"hrs") + " ";
    if(minutes && minutes > 0) out += minutes + " " + ((minutes == 1)?"min":"mins") + " ";
    if(seconds && seconds > 0) out += seconds + " " + ((seconds == 1)?"sec":"secs") + " ";
    if(millis&& millis> 0) out += millis+ " " + ((millis== 1)?"msec":"msecs") + " ";
    return out.trim();
}

2

我建议使用您选择的语言/框架提供的任何日期/时间功能/库。还请检查字符串格式化功能,因为它们通常提供传递日期/时间戳和输出人类可读的字符串格式的简便方法。


2

您的选择很简单:

  1. 编写代码进行转换(即,除以milliSecondsPerDay以获得天数,并使用模数除以milliSecondsPerHour以获得小时,然后使用模数除以milliSecondsPerMinute并除以1000秒钟得到秒数。milliSecondsPerMinute = 60000,milliSecondsPerHour = 60 * milliSecondsPerMinute,milliSecondsPerDay = 24 * milliSecondsPerHour。
  2. 使用某种操作程序。UNIX和Windows都具有可以从Ticks或seconds类型值获得的结构。

2
Long serverUptimeSeconds = 
    (System.currentTimeMillis() - SINCE_TIME_IN_MILLISECONDS) / 1000;


String serverUptimeText = 
String.format("%d days %d hours %d minutes %d seconds",
serverUptimeSeconds / 86400,
( serverUptimeSeconds % 86400) / 3600 ,
((serverUptimeSeconds % 86400) % 3600 ) / 60,
((serverUptimeSeconds % 86400) % 3600 ) % 60
);

2
Long expireTime = 69l;
Long tempParam = 0l;

Long seconds = math.mod(expireTime, 60);
tempParam = expireTime - seconds;
expireTime = tempParam/60;
Long minutes = math.mod(expireTime, 60);
tempParam = expireTime - minutes;
expireTime = expireTime/60;
Long hours = math.mod(expireTime, 24);
tempParam = expireTime - hours;
expireTime = expireTime/24;
Long days = math.mod(expireTime, 30);

system.debug(days + '.' + hours + ':' + minutes + ':' + seconds);

这应该打印:0.0:1:9


2

为什么不做这样的事情:

var ms = 86400;

var seconds = ms / 1000; //86.4

var分钟=秒/ 60; //1.4400000000000002

var hours =分钟/ 60; //0.024000000000000004

var days = hours / 24; //0.0010000000000000002

并处理浮点精度,例如Number(minutes.toFixed(5))//1.44


2

在java中

public static String formatMs(long millis) {
    long hours = TimeUnit.MILLISECONDS.toHours(millis);
    long mins = TimeUnit.MILLISECONDS.toMinutes(millis);
    long secs = TimeUnit.MILLISECONDS.toSeconds(millis);
    return String.format("%dh %d min, %d sec",
            hours,
            mins - TimeUnit.HOURS.toMinutes(hours),
            secs - TimeUnit.MINUTES.toSeconds(mins)
    );
}

给出这样的东西:

12h 1 min, 34 sec

1

我无法评论您的问题的第一个答案,但是有一个小错误。您应该使用parseInt或Math.floor将浮点数转换为整数,即

var days, hours, minutes, seconds, x;
x = ms / 1000;
seconds = Math.floor(x % 60);
x /= 60;
minutes = Math.floor(x % 60);
x /= 60;
hours = Math.floor(x % 24);
x /= 24;
days = Math.floor(x);

就个人而言,我在项目中使用CoffeeScript,我的代码如下所示:

getFormattedTime : (ms)->
        x = ms / 1000
        seconds = Math.floor x % 60
        x /= 60
        minutes = Math.floor x % 60
        x /= 60
        hours = Math.floor x % 24
        x /= 24
        days = Math.floor x
        formattedTime = "#{seconds}s"
        if minutes then formattedTime = "#{minutes}m " + formattedTime
        if hours then formattedTime = "#{hours}h " + formattedTime
        formattedTime 

1

这是一个解决方案。稍后,您可以用“:”分割并获取数组的值

/**
 * Converts milliseconds to human readeable language separated by ":"
 * Example: 190980000 --> 2:05:3 --> 2days 5hours 3min
 */
function dhm(t){
    var cd = 24 * 60 * 60 * 1000,
        ch = 60 * 60 * 1000,
        d = Math.floor(t / cd),
        h = '0' + Math.floor( (t - d * cd) / ch),
        m = '0' + Math.round( (t - d * cd - h * ch) / 60000);
    return [d, h.substr(-2), m.substr(-2)].join(':');
}

var delay = 190980000;                   
var fullTime = dhm(delay);
console.log(fullTime);

1

这是我使用TimeUnit的解决方案。

更新:我应该指出,这是用groovy编写的,但是Java几乎是相同的。

def remainingStr = ""

/* Days */
int days = MILLISECONDS.toDays(remainingTime) as int
remainingStr += (days == 1) ? '1 Day : ' : "${days} Days : "
remainingTime -= DAYS.toMillis(days)

/* Hours */
int hours = MILLISECONDS.toHours(remainingTime) as int
remainingStr += (hours == 1) ? '1 Hour : ' : "${hours} Hours : "
remainingTime -= HOURS.toMillis(hours)

/* Minutes */
int minutes = MILLISECONDS.toMinutes(remainingTime) as int
remainingStr += (minutes == 1) ? '1 Minute : ' : "${minutes} Minutes : "
remainingTime -= MINUTES.toMillis(minutes)

/* Seconds */
int seconds = MILLISECONDS.toSeconds(remainingTime) as int
remainingStr += (seconds == 1) ? '1 Second' : "${seconds} Seconds"

1

一种灵活的方法:(
不适用于当前日期,但足以满足持续时间)

/**
convert duration to a ms/sec/min/hour/day/week array
@param {int}        msTime              : time in milliseconds 
@param {bool}       fillEmpty(optional) : fill array values even when they are 0.
@param {string[]}   suffixes(optional)  : add suffixes to returned values.
                                        values are filled with missings '0'
@return {int[]/string[]} : time values from higher to lower(ms) range.
*/
var msToTimeList=function(msTime,fillEmpty,suffixes){
    suffixes=(suffixes instanceof Array)?suffixes:[];   //suffixes is optional
    var timeSteps=[1000,60,60,24,7];    // time ranges : ms/sec/min/hour/day/week
    timeSteps.push(1000000);    //add very big time at the end to stop cutting
    var result=[];
    for(var i=0;(msTime>0||i<1||fillEmpty)&&i<timeSteps.length;i++){
        var timerange = msTime%timeSteps[i];
        if(typeof(suffixes[i])=="string"){
            timerange+=suffixes[i]; // add suffix (converting )
            // and fill zeros :
            while(  i<timeSteps.length-1 &&
                    timerange.length<((timeSteps[i]-1)+suffixes[i]).length  )
                timerange="0"+timerange;
        }
        result.unshift(timerange);  // stack time range from higher to lower
        msTime = Math.floor(msTime/timeSteps[i]);
    }
    return result;
};

注意:如果要控制时间范围,也可以将timeSteps设置为参数。

使用方法(复制测试):

var elsapsed = Math.floor(Math.random()*3000000000);

console.log(    "elsapsed (labels) = "+
        msToTimeList(elsapsed,false,["ms","sec","min","h","days","weeks"]).join("/")    );

console.log(    "half hour : "+msToTimeList(elsapsed,true)[3]<30?"first":"second"   );

console.log(    "elsapsed (classic) = "+
        msToTimeList(elsapsed,false,["","","","","",""]).join(" : ")    );

1

我建议使用http://www.ocpsoft.org/prettytime/ 库。

以人类可读的形式获取时间间隔非常简单,例如

PrettyTime p = new PrettyTime(); System.out.println(p.format(new Date()));

它会像“从现在起的瞬间”打印

其他例子

PrettyTime p = new PrettyTime()); Date d = new Date(System.currentTimeMillis()); d.setHours(d.getHours() - 1); String ago = p.format(d);

然后字符串ago =“ 1小时前”


0

这是JAVA中更精确的方法,我已经实现了这个简单的逻辑,希望对您有所帮助:

    public String getDuration(String _currentTimemilliSecond)
    {
        long _currentTimeMiles = 1;         
        int x = 0;
        int seconds = 0;
        int minutes = 0;
        int hours = 0;
        int days = 0;
        int month = 0;
        int year = 0;

        try 
        {
            _currentTimeMiles = Long.parseLong(_currentTimemilliSecond);
            /**  x in seconds **/   
            x = (int) (_currentTimeMiles / 1000) ; 
            seconds = x ;

            if(seconds >59)
            {
                minutes = seconds/60 ;

                if(minutes > 59)
                {
                    hours = minutes/60;

                    if(hours > 23)
                    {
                        days = hours/24 ;

                        if(days > 30)
                        {
                            month = days/30;

                            if(month > 11)
                            {
                                year = month/12;

                                Log.d("Year", year);
                                Log.d("Month", month%12);
                                Log.d("Days", days % 30);
                                Log.d("hours ", hours % 24);
                                Log.d("Minutes ", minutes % 60);
                                Log.d("Seconds  ", seconds % 60);   

                                return "Year "+year + " Month "+month%12 +" Days " +days%30 +" hours "+hours%24 +" Minutes "+minutes %60+" Seconds "+seconds%60;
                            }
                            else
                            {
                                Log.d("Month", month);
                                Log.d("Days", days % 30);
                                Log.d("hours ", hours % 24);
                                Log.d("Minutes ", minutes % 60);
                                Log.d("Seconds  ", seconds % 60);   

                                return "Month "+month +" Days " +days%30 +" hours "+hours%24 +" Minutes "+minutes %60+" Seconds "+seconds%60;
                            }

                        }
                        else
                        {
                            Log.d("Days", days );
                            Log.d("hours ", hours % 24);
                            Log.d("Minutes ", minutes % 60);
                            Log.d("Seconds  ", seconds % 60);   

                            return "Days " +days +" hours "+hours%24 +" Minutes "+minutes %60+" Seconds "+seconds%60;
                        }

                    }
                    else
                    {
                        Log.d("hours ", hours);
                        Log.d("Minutes ", minutes % 60);
                        Log.d("Seconds  ", seconds % 60);

                        return "hours "+hours+" Minutes "+minutes %60+" Seconds "+seconds%60;
                    }
                }
                else
                {
                    Log.d("Minutes ", minutes);
                    Log.d("Seconds  ", seconds % 60);

                    return "Minutes "+minutes +" Seconds "+seconds%60;
                }
            }
            else
            {
                Log.d("Seconds ", x);
                return " Seconds "+seconds;
            }
        }
        catch (Exception e) 
        {
            Log.e(getClass().getName().toString(), e.toString());
        }
        return "";
    }

    private Class Log
    {
        public static void d(String tag , int value)
        {
            System.out.println("##### [ Debug ]  ## "+tag +" :: "+value);
        }
    }
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