Javascript返回两个日期之间的天数，小时数，分钟数，秒数-Java 学习之路

有没有人可以链接我的一些教程，我可以找到如何在2个unix日期之间的javascript中返回天，小时，分钟，秒？

我有：

var date_now = unixtimestamp;
var date_future = unixtimestamp;

我想返回（实时）从date_now到date_future的剩余天数，小时数，分钟数 .

11 回答

只需计算出以秒为单位的差异（不要忘记JS时间戳实际上以毫秒为单位）并分解该值：

// get total seconds between the times
var delta = Math.abs(date_future - date_now) / 1000;

// calculate (and subtract) whole days
var days = Math.floor(delta / 86400);
delta -= days * 86400;

// calculate (and subtract) whole hours
var hours = Math.floor(delta / 3600) % 24;
delta -= hours * 3600;

// calculate (and subtract) whole minutes
var minutes = Math.floor(delta / 60) % 60;
delta -= minutes * 60;

// what's left is seconds
var seconds = delta % 60;  // in theory the modulus is not required

EDIT 代码已经调整，因为我刚刚意识到原始代码返回了总小时数等，而不是整天计算后剩余的小时数 .

回复于 2024-04-28T14:03:34+08:00

这里是javascript :(例如，未来的日期是新年）

DEMO （每秒更新）

var dateFuture = new Date(new Date().getFullYear() +1, 0, 1);
var dateNow = new Date();

var seconds = Math.floor((dateFuture - (dateNow))/1000);
var minutes = Math.floor(seconds/60);
var hours = Math.floor(minutes/60);
var days = Math.floor(hours/24);

hours = hours-(days*24);
minutes = minutes-(days*24*60)-(hours*60);
seconds = seconds-(days*24*60*60)-(hours*60*60)-(minutes*60);

回复于 2024-04-28T14:03:34+08:00

我称之为"snowman-carl ☃ method"，我认为当你需要额外的时间 Span ，如周，飞蛾，年，几个世纪时，它会更灵活一点......并且不需要太多的重复代码：

var d = Math.abs(date_future - date_now) / 1000;                           // delta
var r = {};                                                                // result
var s = {                                                                  // structure
    year: 31536000,
    month: 2592000,
    week: 604800, // uncomment row to ignore
    day: 86400,   // feel free to add your own row
    hour: 3600,
    minute: 60,
    second: 1
};

Object.keys(s).forEach(function(key){
    r[key] = Math.floor(d / s[key]);
    d -= r[key] * s[key];
});

// for example: {year:0,month:0,week:1,day:2,hour:34,minute:56,second:7}
console.log(r);

拥有FIDDLE / ES6版本（2018）/ TypeScript版本（2019）

灵感来自Alnitak的回答 .

回复于 2024-04-28T14:03:34+08:00

19

请注意，仅根据差异进行计算不会涵盖所有情况：闰年和“夏令时”的切换 .

Javascript用于处理日期的内置库很差 . 我建议你使用第三方javascript库，例如MomentJS;你可以看到here你正在寻找的功能 .

回复于 2024-04-28T14:03:34+08:00

我知道持续时间细分的最佳库是countdown.js . 它处理所有困难情况，如闰年和夏令时csg mentioned，甚至允许您指定模糊概念，如月和周 . 这是您案例的代码：

//assuming these are in *seconds* (in case of MS don't multiply by 1000 below)
var date_now = 1218374; 
var date_future = 29384744;

diff = countdown(date_now * 1000, date_future * 1000, 
            countdown.DAYS | countdown.HOURS | countdown.MINUTES | countdown.SECONDS);
alert("days: " + diff.days + " hours: " + diff.hours + 
      " minutes: " + diff.minutes + " seconds: " + diff.seconds);

//or even better
alert(diff.toString());

这是一个JSFiddle，但它可能仅适用于禁用Web安全的FireFox或Chrome，因为countdown.js托管文本/纯MIME类型（您应该提供文件，而不是链接到countdownjs.org） .

回复于 2024-04-28T14:03:34+08:00

function update(datetime = "2017-01-01 05:11:58") {
    var theevent = new Date(datetime);
    now = new Date();
    var sec_num = (theevent - now) / 1000;
    var days    = Math.floor(sec_num / (3600 * 24));
    var hours   = Math.floor((sec_num - (days * (3600 * 24)))/3600);
    var minutes = Math.floor((sec_num - (days * (3600 * 24)) - (hours * 3600)) / 60);
    var seconds = Math.floor(sec_num - (days * (3600 * 24)) - (hours * 3600) - (minutes * 60));

    if (hours   < 10) {hours   = "0"+hours;}
    if (minutes < 10) {minutes = "0"+minutes;}
    if (seconds < 10) {seconds = "0"+seconds;}

    return  days+':'+ hours+':'+minutes+':'+seconds;
}

回复于 2024-04-28T14:03:34+08:00

这是一个代码示例 . 我使用简单的计算而不是使用预先计算，如1天是86400秒 . 所以你可以轻松地遵循逻辑 .

// Calculate time between two dates:
var date1 = new Date('1110-01-01 11:10');
var date2 = new Date();

console.log('difference in ms', date1 - date2);

// Use Math.abs() so the order of the dates can be ignored and you won't
// end up with negative numbers when date1 is before date2.
console.log('difference in ms abs', Math.abs(date1 - date2));
console.log('difference in seconds', Math.abs(date1 - date2) / 1000);

var diffInSeconds = Math.abs(date1 - date2) / 1000;
var days = Math.floor(diffInSeconds / 60 / 60 / 24);
var hours = Math.floor(diffInSeconds / 60 / 60 % 24);
var minutes = Math.floor(diffInSeconds / 60 % 60);
var seconds = Math.floor(diffInSeconds % 60);
var milliseconds = Math.round((diffInSeconds - Math.floor(diffInSeconds)) * 1000);

console.log('days', days);
console.log('hours', ('0' + hours).slice(-2));
console.log('minutes', ('0' + minutes).slice(-2));
console.log('seconds', ('0' + seconds).slice(-2));
console.log('milliseconds', ('00' + milliseconds).slice(-3));

回复于 2024-04-28T14:03:34+08:00

使用moment.js库，例如：

var time = date_future - date_now;
var seconds = moment.duration(time).seconds();
var minutes = moment.duration(time).minutes();
var hours   = moment.duration(time).hours();
var days    = moment.duration(time).days();

回复于 2024-04-28T14:03:34+08:00

124

这是一个代码，用于查找天，小时，分钟，秒中两个日期之间的差异（假设未来日期是新年日期） .

var one_day = 24*60*60*1000;              // total milliseconds in one day

var today = new Date();
var new_year = new Date("01/01/2017");    // future date

var today_time = today.getTime();         // time in miliiseconds
var new_year_time = new_year.getTime();                         

var time_diff = Math.abs(new_year_time - today_time);  //time diff in ms  
var days = Math.floor(time_diff / one_day);            // no of days

var remaining_time = time_diff - (days*one_day);      // remaining ms  

var hours = Math.floor(remaining_time/(60*60*1000));   
remaining_time = remaining_time - (hours*60*60*1000);  

var minutes = Math.floor(remaining_time/(60*1000));        
remaining_time = remaining_time - (minutes * 60 * 1000);   

var seconds = Math.ceil(remaining_time / 1000);

回复于 2024-04-28T14:03:34+08:00

let delta = Math.floor(Math.abs(start.getTime() - end.getTime()) / 1000);
let hours = Math.floor(delta / 3600);
delta -= hours * 3600;
let minutes = Math.floor(delta / 60);
delta -= minutes * 60;
let seconds = delta;
if (hours.toString().length === 1) {
  hours = `0${hours}`;
}
if (minutes.toString().length === 1) {
  minutes = `0${minutes}`;
}
if (seconds.toString().length === 1) {
  seconds = `0${seconds}`;
}
const recordingTime = `${hours}:${minutes}:${seconds}`;

回复于 2024-04-28T14:03:34+08:00

Short and flexible with support for negative values ，虽然使用两个逗号表达式:)

function timeUnitsBetween(startDate, endDate) {
  let delta = Math.abs(endDate - startDate) / 1000;
  const isNegative = startDate > endDate ? -1 : 1;
  return [
    ['days', 24 * 60 * 60],
    ['hours', 60 * 60],
    ['minutes', 60],
    ['seconds', 1]
  ].reduce((acc, [key, value]) => (acc[key] = Math.floor(delta / value) * isNegative, delta -= acc[key] * isNegative * value, acc), {});
}

Example:

timeUnitsBetween(new Date("2019-02-11T02:12:03+00:00"), new Date("2019-02-11T01:00:00+00:00"));
// { days: -0, hours: -1, minutes: -12, seconds: -3 }

受RienNeVaPlu͢s解决方案的启发 .

回复于 2024-04-28T14:03:34+08:00

Javascript返回两个日期之间的天数，小时数，分钟数，秒数

11 回答

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