我需要将任意数量的毫秒转换为天、小时、分秒。
例如：10 天、5 小时、13 分钟、1 秒。
Jonathan Holland
问问题
158002 次
22 回答
230
好吧，既然没有其他人加紧，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
我很高兴你几天都停下来，几个月都没有问。:)
注意，在上面，假设/表示截断整数除法。/如果您在表示浮点除法的语言中使用此代码，则需要根据需要手动截断除法的结果。
于 2008-10-06T18:28:43.877 回答
60
令 A 为毫秒数。然后你有：
seconds=(A/1000)%60
minutes=(A/(1000*60))%60
hours=(A/(1000*60*60))%24
依此类推（%是模运算符）。
希望这可以帮助。
于 2008-10-06T18:28:42.877 回答
28
以下两种解决方案都使用javascript（我不知道该解决方案与语言无关！）。如果捕获持续时间，这两种解决方案都需要扩展> 1 month。
解决方案 1：使用 Date 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！:)
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;
};
使用您需要的任何字段创建一个“持续时间”对象。格式化时间戳然后变得简单......
console.log(getDuration(536643021).toString());
给出：
"6 days, 5 hours, 4 minutes, 3 seconds, 21 millis"
于 2012-09-14T08:29:25.300 回答
23
Apache Commons Lang有一个DurationFormatUtils，它具有非常有用的方法，例如formatDurationWords。
于 2011-11-09T18:08:44.310 回答
8
您应该使用您正在使用的任何语言的日期时间函数，但是，只是为了好玩，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;
于 2008-10-06T18:29:35.310 回答
4
这是我写的一个方法。它需要一个integer milliseconds value并返回一个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;
}
于 2012-01-09T21:33:09.927 回答
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();
}
于 2012-10-31T05:39:41.647 回答
3
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
于 2016-08-13T16:52:08.553 回答
2
我建议使用您选择的语言/框架提供的任何日期/时间函数/库。还要查看字符串格式化函数，因为它们通常提供简单的方法来传递日期/时间戳并输出人类可读的字符串格式。
于 2008-10-06T18:25:37.503 回答
2
您的选择很简单：
- 编写代码进行转换（即，除以milliSecondsPerDay 得到天数，使用模数除以milliSecondsPerHour 得到小时数，使用模数除以milliSecondsPerMinute 并除以1000 得到秒数。milliSecondsPerMinute = 60000, milliSecondsPerHour = 60 * milliSecondsPerMinute，milliSecondsPerDay = 24 * milliSecondsPerHour。
- 使用某种操作程序。UNIX 和 Windows 都具有可以从 Ticks 或 seconds 类型值中获取的结构。
于 2008-10-06T18:29:13.627 回答
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
);
于 2010-10-15T15:23:51.043 回答
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
于 2013-06-27T20:35:00.027 回答
2
为什么不做这样的事情：
无功毫秒 = 86400;
var 秒 = 毫秒 / 1000；//86.4
var 分钟 = 秒 / 60；//1.4400000000000002
var 小时 = 分钟 / 60；//0.024000000000000004
var 天 = 小时 / 24；//0.0010000000000000002
并处理浮点精度，例如 Number(minutes.toFixed(5)) //1.44
于 2015-01-21T10:47:50.097 回答
1
我无法评论您的问题的第一个答案，但有一个小错误。您应该使用 parseInt 或 Math.floor 将浮点数转换为整数，我
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，我在我的项目中使用 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
于 2012-11-14T10:24:43.350 回答
1
这是一个解决方案。稍后您可以按“：”拆分并获取数组的值
/**
* 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);
于 2013-05-23T13:43:31.323 回答
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 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"
于 2014-07-04T14:20:29.950 回答
1
一种灵活的方法：（
不适用于当前日期，但对于持续时间足够好）
/**
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：如果你想控制时间范围， 你也可以将timeSteps设置为参数。
如何使用（复制一个测试）：
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(" : ") );
于 2015-01-17T20:43:33.143 回答
1
我建议使用http://www.ocpsoft.org/prettytime/ library..
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“从现在开始”一样打印
另一个例子
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"
于 2015-03-21T10:35:46.407 回答
1
在 python 3 中，您可以使用以下代码段来实现您的目标：
from datetime import timedelta
ms = 536643021
td = timedelta(milliseconds=ms)
print(str(td))
# --> 6 days, 5:04:03.021000
时间增量文档：https ://docs.python.org/3/library/datetime.html#datetime.timedelta
timedelta str的__str__方法来源：https ://github.com/python/cpython/blob/33922cb0aa0c81ebff91ab4e938a58dfec2acf19/Lib/datetime.py#L607
于 2020-11-12T08:08:57.183 回答
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);
}
}
于 2012-10-01T07:04:26.863 回答
0
使用的解决方案awk：
$ ms=10000001; awk -v ms=$ms 'BEGIN {x=ms/1000;
s=x%60; x/=60;
m=x%60; x/=60;
h=x%60;
printf("%02d:%02d:%02d.%03d\n", h, m, s, ms%1000)}'
02:46:40.001
于 2020-11-06T20:37:36.673 回答
0
这个遗漏了 0 个值。带测试。
const toTimeString = (value, singularName) =>
`${value} ${singularName}${value !== 1 ? 's' : ''}`;
const readableTime = (ms) => {
const days = Math.floor(ms / (24 * 60 * 60 * 1000));
const daysMs = ms % (24 * 60 * 60 * 1000);
const hours = Math.floor(daysMs / (60 * 60 * 1000));
const hoursMs = ms % (60 * 60 * 1000);
const minutes = Math.floor(hoursMs / (60 * 1000));
const minutesMs = ms % (60 * 1000);
const seconds = Math.round(minutesMs / 1000);
const data = [
[days, 'day'],
[hours, 'hour'],
[minutes, 'minute'],
[seconds, 'second'],
];
return data
.filter(([value]) => value > 0)
.map(([value, name]) => toTimeString(value, name))
.join(', ');
};
// Tests
const hundredDaysTwentyHoursFiftyMinutesThirtySeconds = 8715030000;
const oneDayTwoHoursEightMinutesTwelveSeconds = 94092000;
const twoHoursFiftyMinutes = 10200000;
const oneMinute = 60000;
const fortySeconds = 40000;
const oneSecond = 1000;
const oneDayTwelveSeconds = 86412000;
const test = (result, expected) => {
console.log(expected, '- ' + (result === expected));
};
test(readableTime(
hundredDaysTwentyHoursFiftyMinutesThirtySeconds
), '100 days, 20 hours, 50 minutes, 30 seconds');
test(readableTime(
oneDayTwoHoursEightMinutesTwelveSeconds
), '1 day, 2 hours, 8 minutes, 12 seconds');
test(readableTime(
twoHoursFiftyMinutes
), '2 hours, 50 minutes');
test(readableTime(
oneMinute
), '1 minute');
test(readableTime(
fortySeconds
), '40 seconds');
test(readableTime(
oneSecond
), '1 second');
test(readableTime(
oneDayTwelveSeconds
), '1 day, 12 seconds');于 2021-08-30T17:17:37.713 回答