有没有人可以将我链接到一些教程，我可以在其中找到如何在 2 个 unix 日期时间之间返回 javascript 中的天数、小时数、分钟数、秒数？
我有：
var date_now = unixtimestamp;
var date_future = unixtimestamp;
我想返回（实时）从 date_now 到 date_future 还剩多少天、小时、分钟、秒。
问问题
191356 次
22 回答
215
只需计算以秒为单位的差异（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
编辑代码已调整，因为我刚刚意识到原始代码返回的总小时数等，而不是计算一整天后剩下的小时数。
于 2012-12-16T18:13:08.363 回答
53
这是javascript中的：（例如，未来的日期是元旦）
演示（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);
于 2012-12-16T19:13:51.773 回答
45
我称之为“雪人 ☃ 方法”，我认为当您需要额外的时间跨度（如几周、飞蛾、年、世纪……）并且不想要太多重复的代码时，它会更加灵活：
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);
有 一个FIDDLE//ES6 Version (2018)TypeScript Version (2019)
灵感来自Alnitak的回答。
于 2015-09-11T01:42:15.240 回答
10
我的解决方案不是那么清楚，但我把它作为另一个例子
console.log(duration('2019-07-17T18:35:25.235Z', '2019-07-20T00:37:28.839Z'));
function duration(t0, t1){
let d = (new Date(t1)) - (new Date(t0));
let weekdays = Math.floor(d/1000/60/60/24/7);
let days = Math.floor(d/1000/60/60/24 - weekdays*7);
let hours = Math.floor(d/1000/60/60 - weekdays*7*24 - days*24);
let minutes = Math.floor(d/1000/60 - weekdays*7*24*60 - days*24*60 - hours*60);
let seconds = Math.floor(d/1000 - weekdays*7*24*60*60 - days*24*60*60 - hours*60*60 - minutes*60);
let milliseconds = Math.floor(d - weekdays*7*24*60*60*1000 - days*24*60*60*1000 - hours*60*60*1000 - minutes*60*1000 - seconds*1000);
let t = {};
['weekdays', 'days', 'hours', 'minutes', 'seconds', 'milliseconds'].forEach(q=>{ if (eval(q)>0) { t[q] = eval(q); } });
return t;
}于 2019-07-20T00:08:01.020 回答
8
有点不同的风格（也许对于某些人来说更具可读性）它可以在 JavaScript 中使用，而且它也可以在 TypeScript中使用。
如果您确保第一个日期始终大于第二个日期，则不需要 Math.abs() 模运算周围的圆括号也是不必要的。我保留它们以备清关。
let diffTime = Math.abs(new Date().valueOf() - new Date('2021-11-22T18:30:00').valueOf());
let days = diffTime / (24*60*60*1000);
let hours = (days % 1) * 24;
let minutes = (hours % 1) * 60;
let secs = (minutes % 1) * 60;
[days, hours, minutes, secs] = [Math.floor(days), Math.floor(hours), Math.floor(minutes), Math.floor(secs)]
console.log(days+'d', hours+'h', minutes+'m', secs+'s');于 2021-10-04T02:15:55.610 回答
6
这是一个代码示例。我使用了简单的计算，而不是使用像 1 天是 86400 秒这样的预先计算。因此，您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));
于 2016-09-20T09:50:09.577 回答
5
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();
于 2018-09-28T07:15:56.427 回答
5
简短而灵活，支持负值，尽管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), {});
}
例子：
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 }
受到RienNeVaPlu͢解决方案的启发。
于 2019-02-10T12:33:29.737 回答
4
我所知道的关于持续时间细分的最佳库是countdown.js。它处理所有困难的情况，如csg 提到的闰年和夏令时，甚至允许您指定模糊概念，如月和周。这是您的案例的代码：
//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，但它可能只能在 FireFox 或禁用网络安全的 Chrome 中工作，因为 countdown.js 以text/plain MIME 类型托管（您应该提供文件，而不是链接到 countdownjs.org） .
于 2015-02-15T08:27:14.773 回答
4
对于那些只想要小时和分钟的人来说，使用这个
const oldDate = new Date("2021-04-28T13:17:31.000Z")
const newDate = new Date("2021-04-28T22:08:07.000Z")
const msToTime = (ms) => ({
hours: Math.trunc(ms/3600000),
minutes: Math.trunc((ms/3600000 - Math.trunc(ms/3600000))*60) + ((ms/3600000 - Math.trunc(ms/3600000))*60 % 1 != 0 ? 1 : 0)
})
console.log(msToTime(Math.abs(newDate-oldDate)))于 2021-04-30T02:13:44.740 回答
3
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;
}
于 2016-07-02T07:21:08.053 回答
3
因为 MomentJS 相当重且未优化，所以不怕使用模块的人可能应该看看date-fns，它提供了一个intervalToDuration方法，它可以满足你的需求：
const result = intervalToDuration({
start: new Date(dateNow),
end: new Date(dateFuture),
})
这将返回一个看起来像这样的对象：
{
years: 39,
months: 2,
days: 20,
hours: 7,
minutes: 5,
seconds: 0,
}
然后您甚至可以使用您喜欢的参数使用formatDuration将此对象显示为字符串
于 2021-02-02T02:27:05.810 回答
2
MomentJS 有一个功能可以做到这一点：
const start = moment(j.timings.start);
const end = moment(j.timings.end);
const elapsedMinutes = end.diff(start, "minutes");
于 2019-05-13T13:55:35.687 回答
2
这是我的看法：
timeSince(123456) => "1 day, 10 hours, 17 minutes, 36 seconds"
和代码：
function timeSince(date, longText) {
let seconds = null;
let leadingText = null;
if (date instanceof Date) {
seconds = Math.floor((new Date() - date) / 1000);
if (seconds < 0) {
leadingText = " from now";
} else {
leadingText = " ago";
}
seconds = Math.abs(seconds);
} else {
seconds = date;
leadingText = "";
}
const intervals = [
[31536000, "year" ],
[ 2592000, "month" ],
[ 86400, "day" ],
[ 3600, "hour" ],
[ 60, "minute"],
[ 1, "second"],
];
let interval = seconds;
let intervalStrings = [];
for (let i = 0; i < intervals.length; i++) {
let divResult = Math.floor(interval / intervals[i][0]);
if (divResult > 0) {
intervalStrings.push(divResult + " " + intervals[i][1] + ((divResult > 1) ? "s" : ""));
interval = interval % intervals[i][0];
if (!longText) {
break;
}
}
}
let intStr = intervalStrings.join(", ");
return intStr + leadingText;
}
于 2019-10-28T11:29:16.000 回答
2
简单的方法
function diff_hours(dt2, dt1)
{
var diff =(dt2.getTime() - dt1.getTime()) / 1000;
diff /= (60 * 60);
return Math.abs(Math.round(diff));
}
function diff_minutes(dt2, dt1)
{
var diff =(dt2.getTime() - dt1.getTime()) / 1000;
diff /= (60);
return Math.abs(Math.round(diff));
}
function diff_seconds(dt2, dt1)
{
var diff =(dt2.getTime() - dt1.getTime()) / 1000;
return Math.abs(Math.round(diff));
}
function diff_miliseconds(dt2, dt1)
{
var diff =(dt2.getTime() - dt1.getTime());
return Math.abs(Math.round(diff));
}
dt1 = new Date(2014,10,2);
dt2 = new Date(2014,10,3);
console.log(diff_hours(dt1, dt2));
dt1 = new Date("October 13, 2014 08:11:00");
dt2 = new Date("October 14, 2014 11:13:00");
console.log(diff_hours(dt1, dt2));
console.log(diff_minutes(dt1, dt2));
console.log(diff_seconds(dt1, dt2));
console.log(diff_miliseconds(dt1, dt2));
于 2020-01-24T13:45:44.903 回答
1
function calculateExamRemainingTime(exam_end_at) {
$(function(){
const calcNewYear = setInterval(function(){
const exam_ending_at = new Date(exam_end_at);
const current_time = new Date();
const totalSeconds = Math.floor((exam_ending_at - (current_time))/1000);;
const totalMinutes = Math.floor(totalSeconds/60);
const totalHours = Math.floor(totalMinutes/60);
const totalDays = Math.floor(totalHours/24);
const hours = totalHours - ( totalDays * 24 );
const minutes = totalMinutes - ( totalDays * 24 * 60 ) - ( hours * 60 );
const seconds = totalSeconds - ( totalDays * 24 * 60 * 60 ) - ( hours * 60 * 60 ) - ( minutes * 60 );
const examRemainingHoursSection = document.querySelector('#remainingHours');
const examRemainingMinutesSection = document.querySelector('#remainingMinutes');
const examRemainingSecondsSection = document.querySelector('#remainingSeconds');
examRemainingHoursSection.innerHTML = hours.toString();
examRemainingMinutesSection.innerHTML = minutes.toString();
examRemainingSecondsSection.innerHTML = seconds.toString();
},1000);
});
}
calculateExamRemainingTime('2025-06-03 20:20:20');
于 2020-06-28T14:25:25.253 回答
1
const arrDiff = [
{
label: 'second',
value: 1000,
},
{
label: 'minute',
value: 1000 * 60,
},
{
label: 'hour',
value: 1000 * 60 * 60,
},
{
label: 'day',
value: 1000 * 60 * 60 * 24,
},
{
label: 'year',
value: 1000 * 60 * 60 * 24 * 365,
},
];
function diff(date) {
let result = { label: '', value: '' };
const value = Date.now() - date;
for (let obj of arrDiff) {
let temp = Math.round(value / obj.value);
if (temp === 0) {
break;
} else
result = {
value: temp,
label: obj.label,
};
}
return result;
}
const postDate = new Date('2020-12-17 23:50:00+0700');
console.log(diff(postDate));
于 2020-12-17T17:09:54.747 回答
0
这是一个代码，用于查找天、小时、分钟、秒中的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);
于 2017-01-19T17:58:27.490 回答
0
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}`;
于 2018-12-23T10:04:28.943 回答
0
我们可以通过简单的方法做到
/*Declare the function */
function Clock(){
let d1 = new Date("1 Jan 2021");
let d2 = new Date();
let difference = Math.abs(d1 - d2); //to get absolute value
//calculate for each one
let Days = Math.floor(difference / ( 1000 * 60 * 60 * 24 ));
let Hours = Math.floor((difference / ( 1000 * 60 * 60 )) % 24);
let Mins = Math.floor((difference / ( 1000 * 60 )) % 60);
let Seconds = Math.floor((difference / ( 1000 )) % 60);
//getting nodes and change the text inside
let getday = document.querySelector(".big_text_days");
let gethour = document.querySelector(".big_text_hours");
let getmins = document.querySelector(".big_text_mins");
let getsec = document.querySelector(".big_text_sec");
getday.textContent = Check_Zero(Days);
gethour.textContent = Check_Zero(Hours);
getmins.textContent = Check_Zero(Mins)
getsec.textContent = Check_Zero(Seconds);
}
//call the funcion for every 1 second
setInterval(Clock , 1000);
//check and add zero in front, if it is lessthan 10
function Check_Zero(mytime){
return mytime < 10 ? "0"+mytime : mytime;
}*{
padding: 0px;
margin: 0px;
box-sizing: border-box;
}
body{
max-width: 900px;
margin: 0px auto;
background-color:whitesmoke;
background-size: cover;
display: flex;
flex-direction: column;
align-items: center;
margin-top: 5rem;
}
.main_container{
display: flex;
flex-wrap: wrap;
justify-content: center;
}
h1{
font-size: 3rem;
color: #3D4B72;
}
.big_text_days , .big_text_hours , .big_text_mins , .big_text_sec{
font-size: 2rem;
font-weight: bold;
line-height: 2;
color: #AC7591;
text-align: center;
}
p{
padding: 20px 0px 20px 0px;
font-size: 3rem;
text-align: center;
}
.spantext{
color: #103c28;
margin: 0px 3rem;
font-size: 2rem;
font-style: italic;
}
.text_sec{
color : #005259;
}<!DOCTYPE html>
<html>
<head>
<meta charset="UTF-8">
<meta name="description" content="Responsive site">
<meta name="keywords" content="HTML,CSS,JS">
<meta name="author" content="Ranjan">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<title>Home</title>
<link href="https://fonts.googleapis.com/css?family=Alfa+Slab+One|Bree+Serif|Exo|Exo+2|Lato|Mansalva|Playfair+Display&display=swap" rel="stylesheet">
<link rel="stylesheet" href="https://use.fontawesome.com/releases/v5.2.0/css/all.css" integrity="sha384-hWVjflwFxL6sNzntih27bfxkr27PmbbK/iSvJ+a4+0owXq79v+lsFkW54bOGbiDQ" crossorigin="anonymous">
</head>
<body>
<section>
<h1>CountDown Timer</h1>
</section>
<section>
<div class="main_container">
<div class="days_container">
<p class="big_text_days">1</p>
<span class="spantext spantextdays">Days</span>
</div>
<div class="hours_container">
<p class="big_text_hours">1</p>
<span class="spantext spantexthours">Hours</span>
</div>
<div class="mins_container">
<p class="big_text_mins">1</p>
<span class="spantext spantextmins">Minutes</span>
</div>
<div class="sec_container">
<p class="big_text_sec text_sec">1</p>
<span class="spantext spantextsec">Seconds</span>
</div>
</div>
</section>
</body>
</html>于 2020-09-23T16:20:36.067 回答
0
2021版
灵感来自Alnitak和RienNeVaPlus 的答案。
自动计算开始和结束（参见下面的示例）。
const timeUnits = {
year: 31536e6,
month: 2592e6,
week: 6048e5,
day: 864e5,
hour: 36e5,
minute: 6e4,
second: 1e3,
};
const timeUnitsNamesShort = {
year: 'y',
month: 'm',
week: 'w',
day: 'd',
hour: 'h',
minute: 'm',
second: 's',
};
const isFuture = (date) => date > Date.now();
const dateDiffStructure = (date, units = timeUnits) => {
let delta = Math.abs(date - Date.now());
return Object.entries(units).reduce((acc, [key, value]) => {
acc[key] = Math.floor(delta / value);
delta -= acc[key] * value;
return acc;
}, {});
};
const dateDiffStructureToString = (date, diffStructure, values, short) => {
const diffStructureEntries = values
? Object.entries(diffStructure).filter(([key, value]) => values.includes(key) && value)
: Object.entries(diffStructure).filter(([, value]) => value);
if (!diffStructureEntries.length) return 'just now';
const suffix = isFuture(date) ? 'left' : 'ago';
const diffString = diffStructureEntries.reduce((acc, [key, value]) => {
const timeUnit = short
? timeUnitsNamesShort[key]
: value > 1
? ` ${key}s`
: ` ${key}`;
acc = acc ? `${acc} ` : '';
return `${acc}${value}${timeUnit}`;
}, '');
return `${diffString} ${suffix}`;
};
const getDateDiff = (date, values, short) => {
const diffStructure = dateDiffStructure(date);
return dateDiffStructureToString(date, diffStructure, values, short);
};
测试和示例：
const timeUnits = {
year: 31536e6,
month: 2592e6,
week: 6048e5,
day: 864e5,
hour: 36e5,
minute: 6e4,
second: 1e3,
};
const timeUnitsNamesShort = {
year: 'y',
month: 'm',
week: 'w',
day: 'd',
hour: 'h',
minute: 'm',
second: 's',
};
const isFuture = (date) => date > Date.now();
const dateDiffStructure = (date, units = timeUnits) => {
let delta = Math.abs(date - Date.now());
return Object.entries(units).reduce((acc, [key, value]) => {
acc[key] = Math.floor(delta / value);
delta -= acc[key] * value;
return acc;
}, {});
};
const dateDiffStructureToString = (date, diffStructure, values, short) => {
const diffStructureEntries = values ?
Object.entries(diffStructure).filter(([key, value]) => values.includes(key) && value) :
Object.entries(diffStructure).filter(([, value]) => value);
if (!diffStructureEntries.length) return 'just now';
const suffix = isFuture(date) ? 'left' : 'ago';
const diffString = diffStructureEntries.reduce((acc, [key, value]) => {
const timeUnit = short ?
timeUnitsNamesShort[key] :
value > 1 ?
` ${key}s` :
` ${key}`;
acc = acc ? `${acc} ` : '';
return `${acc}${value}${timeUnit}`;
}, '');
return `${diffString} ${suffix}`;
};
const getDateDiff = (date, values, short) => {
const diffStructure = dateDiffStructure(date);
console.log(`dateDiffStructure(${JSON.stringify(date)}) //`, diffStructure);
return dateDiffStructureToString(date, diffStructure, values, short);
};
const tests = [
['After tomorrow', [new Date(Date.now() + 8.64e7 + 1e6)]],
['Yesterday', [new Date(Date.now() - 8.64e7)]],
['Past', [new Date(Date.now() * Math.random())]],
['Future short', [new Date(Date.now() * (Math.random() + 1)), ['year', 'month', 'day'], true]],
['Now', [Date.now()]]
];
tests.forEach(([text, args]) => {
console.log(`${text}:`);
console.group();
console.log(`getDateDiff(${args.map(item => JSON.stringify(item)).join(', ')}) //`, getDateDiff(...args));
console.groupEnd();
});于 2021-03-12T17:48:48.647 回答