如果我有两个日期,如何使用JavaScript在分钟内得到两个日期之间的差异?
当前回答
一个简单的函数来执行这个计算:
function getMinutesBetweenDates(startDate, endDate) {
var diff = endDate.getTime() - startDate.getTime();
return (diff / 60000);
}
其他回答
对于那些喜欢处理小数字的人
const today = new Date();
const endDate = new Date(startDate.setDate(startDate.getDate() + 7));
const days = parseInt((endDate - today) / (1000 * 60 * 60 * 24));
const hours = parseInt(Math.abs(endDate - today) / (1000 * 60 * 60) % 24);
const minutes = parseInt(Math.abs(endDate.getTime() - today.getTime()) / (1000 * 60) % 60);
const seconds = parseInt(Math.abs(endDate.getTime() - today.getTime()) / (1000) % 60);
这应该以分钟为单位显示两个日期之间的差异。在浏览器中试试:
const currDate = new Date('Tue Feb 13 2018 13:04:58 GMT+0200 (EET)')
const oldDate = new Date('Tue Feb 13 2018 12:00:58 GMT+0200 (EET)')
(currDate - oldDate) / 60000 // 64
下面的代码对我有用,
function timeDiffCalc(dateNow,dateFuture) {
var newYear1 = new Date(dateNow);
var newYear2 = new Date(dateFuture);
var dif = (newYear2 - newYear1);
var dif = Math.round((dif/1000)/60);
console.log(dif);
}
这是可行的
duration = moment.duration(moment(end_time).diff(moment(start_time)))
这是我在节点中求解类似问题时得到的一些乐趣。
function formatTimeDiff(date1, date2) {
return Array(3)
.fill([3600, date1.getTime() - date2.getTime()])
.map((v, i, a) => {
a[i+1] = [a[i][0]/60, ((v[1] / (v[0] * 1000)) % 1) * (v[0] * 1000)];
return `0${Math.floor(v[1] / (v[0] * 1000))}`.slice(-2);
}).join(':');
}
const millis = 1000;
const utcEnd = new Date(1541424202 * millis);
const utcStart = new Date(1541389579 * millis);
const utcDiff = formatTimeDiff(utcEnd, utcStart);
console.log(`Dates:
Start : ${utcStart}
Stop : ${utcEnd}
Elapsed : ${utcDiff}
`);
/*
Outputs:
Dates:
Start : Mon Nov 05 2018 03:46:19 GMT+0000 (UTC)
Stop : Mon Nov 05 2018 13:23:22 GMT+0000 (UTC)
Elapsed : 09:37:02
*/
你可以在https://repl.it/@GioCirque/TimeSpan-Formatting上看到它的运行