Question

因此,我有两个日期<代码>。缩略语

我怎么能发现它们之间有多少次?

Answer 1

I m 计及YYYY & ZZZZZZZ系指年、MM &等值的惯性价值;NN系指年月和DD &值;EE作为惯性值,即月日。

var t1 = new Date(YYYY, MM, DD, 0, 0, 0, 0);
var t2 = new Date(ZZZZ, NN, EE, 0, 0, 0, 0);
var dif = t1.getTime() - t2.getTime();

var Seconds_from_T1_to_T2 = dif / 1000;
var Seconds_Between_Dates = Math.abs(Seconds_from_T1_to_T2);

未来的参考文献来源是MDN网站。

或者,如果你的日期以javascript格式出现,那么你可以 par。

var dif = Date.parse(MM + " " + DD + ", " + YYYY) - Date.parse(NN + " " + EE + ", " + ZZZZ);

并且,你可以把这一价值作为微秒的区别(如果我的两个例子都具有同样的含义)

Answer 2

简略:

var a = new Date();
alert("Wait a few seconds, then click OK");

var b = new Date();
var difference = (b - a) / 1000;

console.log("You waited: " + difference + " seconds");

Answer 3

你们只能这样做。

var secondBetweenTwoDate = Math.abs((new Date().getTime() - oldDate.getTime()) / 1000);

Answer 4

如果您的某一个或两个日期是将来的,那么,如果你想要达到第2个准确性,那么我恐怕你会重新使用。 UTC的时间比以前大约6个月大了,因此,任何日期都比某些秒钟可能不准确(实际上,由于人们不更新其机器,你可能发现,今后任何时间都剩下几秒钟)。

这很好地解释了设计日期/时间图书馆的理论,并解释了为什么如此:

Answer 5

var a = new Date("2010 jan 10"),
    b = new Date("2010 jan 9");

alert(
    a + "
" + 
    b + "
" +
    "Difference: " + ((+a - +b) / 1000)
);

Answer 6

这两条都设立了两个<代码>Date 的物体和 ValueOf(),然后对它们进行比较。

Answer 7

2. 便利的方式:

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));

Answer 8

function parseDate(str) {
  const [dateStr, timeStr] = str.split(   );
  const [m,d,y] = dateStr.split( / );
  const [h,min] = timeStr.split( : );
  const date = new Date(y,m,d,h,min,0, 0);
  
  return date;
}

const date1 = parseDate( 28/6/2022 22:55 );
const date2 = parseDate( 28/6/2022 22:58 );
const diffInSeconds = (date2 - date1) / 1000;

console.log(diffInSeconds)

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