分钟持续差分javascript

Minute duration differentiation javascript

本文关键字：javascript 分钟      更新时间：2023-09-26

假设变量date的格式为hour: minute: second 

var time1 = "12:34:19 PM"
var time2 = "12:29:25 PM"

我如何从time1和time2得到微分(持续时间)?我们如何将其转换为日期格式来进行不同的处理?

您可以将时间转换为new Date()，只需在它之前添加一个日期和

使用getTime()获取它们的时间戳并减去它们

var duration = new Date('datetime1').getTime() - new Date('datetime2').getTime()

因为时间戳是结果总秒数的1000倍除以1000

var duration = durantion/1000;

，然后我创建了一个函数来正确格式化秒，使它看起来像一个有效的持续时间

var time1 = "2016-11-02 12:34:19 PM"
var time2 = "2016-11-02 12:29:25 PM"
time1 = new Date(time1 ).getTime();
time2 = new Date(time2 ).getTime();
var duration = (time1 - time2) / 1000;
function formatTime(seconds) {
  var minutes = Math.floor(((seconds/3600)%1)*60);
  minutes = (minutes < 10) ? '0'+minutes : minutes;
  var seconds = Math.round(((seconds/60)%1)*60);
  seconds = (seconds < 10) ? '0'+seconds : seconds;
  return minutes+':'+seconds;
}
console.log('Duration: ' + formatTime(duration)+' secs')

这是工作脚本

String.prototype.toHHMMSS = function () {
    var sec_num = parseInt(this, 10); // don't forget the second param
    var hours   = Math.floor(sec_num / 3600);
    var minutes = Math.floor((sec_num - (hours * 3600)) / 60);
    var seconds = sec_num - (hours * 3600) - (minutes * 60);
    if (hours   < 10) {hours   = "0"+hours;}
    if (minutes < 10) {minutes = "0"+minutes;}
    if (seconds < 10) {seconds = "0"+seconds;}
    return hours+':'+minutes+':'+seconds;
}
var timeStart = new Date("Mon Jan 01 2007 11:00:00 GMT+0530").getTime();
var timeEnd = new Date("Mon Jan 01 2007 11:32:51 GMT+0530").getTime();
var hourDiff = timeEnd - timeStart; //in ms
var secDiff = (hourDiff / 1000).toString(); //in s
alert(secDiff .toHHMMSS());

你可以用这样的构造函数创建一个Date对象:Date(year, month, date, hours, minutes, seconds, ms)

在你的例子中，你可以得到diff:

var date = new Date(2000, 1, 1, 12, 34, 19, 0, 0),
    date2 = new Date(2000, 1, 1, 12, 39, 25, 0, 0);
console.log(date2 - date); // in ms. You can do whatever you want with this value

您可以在这里阅读更多关于Date对象的信息

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