javascript - 在使用 Javascript 确定两个日期之间的差异时，如何考虑闰年？

Question

我正在尝试设置一个计数器来确定设置的过去日期和当前日期之间的差异。

我设法设置了一个计数器来确定两点之间的秒数，使用以下代码将结果分成年、日、小时、分钟、秒：

var lastDay = new Date("Jan 1, 1994  00:00:01").getTime(); 
var x = setInterval(function() { 
var now = new Date().getTime(); 
var t = now - lastDay; 
var years = Math.floor(t / (1000 * 60 * 60 * 24)/ 365);
var days = Math.floor((t % (1000 * 60 * 60 * 24 * 365))/(1000 * 60 * 60 * 24)); 
var hours = Math.floor((t % (1000 * 60 * 60 * 24))/(1000 * 60 * 60)); 
var minutes = Math.floor((t % (1000 * 60 * 60)) / (1000 * 60)); 
var seconds = Math.floor((t % (1000 * 60)) / 1000); 
document.getElementById("year").innerHTML =years ; 
document.getElementById("day").innerHTML =days ; 
document.getElementById("hour").innerHTML =hours; 
document.getElementById("minute").innerHTML = minutes;  
document.getElementById("second").innerHTML =seconds;  
if (t < 0) { 
        clearInterval(x); 
        document.getElementById("demo").innerHTML = "TIME UP"; 
        document.getElementById("year").innerHTML ='0'; 
        document.getElementById("day").innerHTML ='0'; 
        document.getElementById("hour").innerHTML ='0'; 
        document.getElementById("minute").innerHTML ='0' ;  
        document.getElementById("second").innerHTML = '0'; } 
}, 1000);

我遇到的问题是它没有考虑闰年，因此“天”数字不准确。它应该再增加 7 天，以说明设定日期和当前日期（在撰写本文时）之间的闰年数。

我尝试使用下面的代码来计算闰年：

var countLeapYears = function(){
  var yearNow = new Date().getFullYear(); 
    var then = new Date("Jan 1, 1994 00:00:01");
    var yearThen = then.getFullYear();
    var beginYear = 0;
    var endYear = 0;
    var leapYearCount = 0;

    var isLeapYear = function(year){
      return ((year % 4 === 0) && (year % 100 !== 0)) || (year % 400 === 0);
    }

    if(yearNow < y){
       beginYear = yearNow;
       endYear = yearThen;
    }else if(yearNow > yearThen){
       beginYear = yearThen;
       endYear = yearNow;
    }else if(yearNow == yearThen){
       beginYear = yearThen;
       endYear = yearThen;
    }

    for(i = beginYear; i <= endYear; i++){
      if(isLeapYear(i)){
        leapYearCount++;
      }
    }

    return leapYearCount;
}

然后我尝试将 'leapYearCount' 添加到 'days' 但它失败了：

var countLeapYears = function(){
  var yearNow = new Date().getFullYear(); 
    var then = new Date("Jan 1, 1994 00:00:01");
    var yearThen = then.getFullYear();
    var beginYear = 0;
    var endYear = 0;
    var leapYearCount = 0;

    var isLeapYear = function(year){
      return ((year % 4 === 0) && (year % 100 !== 0)) || (year % 400 === 0);
    }

    if(yearNow < y){
       beginYear = yearNow;
       endYear = yearThen;
    }else if(yearNow > yearThen){
       beginYear = yearThen;
       endYear = yearNow;
    }else if(yearNow == yearThen){
       beginYear = yearThen;
       endYear = yearThen;
    }

    for(i = beginYear; i <= endYear; i++){
      if(isLeapYear(i)){
        leapYearCount++;
      }
    }

    return leapYearCount;
}

var lastDay = new Date("Jan 1, 1994 00:00:01").getTime(); 
var x = setInterval(function() { 
var now = new Date().getTime(); 
var t = now - lastDay; 
var years = Math.floor(t / (1000 * 60 * 60 * 24)/ 365);
var days = Math.floor((t % (1000 * 60 * 60 * 24 * 365))/(1000 * 60 * 60 * 24) + leapYearCount); 
var hours = Math.floor((t % (1000 * 60 * 60 * 24))/(1000 * 60 * 60)); 
var minutes = Math.floor((t % (1000 * 60 * 60)) / (1000 * 60)); 
var seconds = Math.floor((t % (1000 * 60)) / 1000); 
document.getElementById("year").innerHTML =years ; 
document.getElementById("day").innerHTML =days ; 
document.getElementById("hour").innerHTML =hours; 
document.getElementById("minute").innerHTML = minutes;  
document.getElementById("second").innerHTML =seconds;  
if (t < 0) { 
        clearInterval(x); 
        document.getElementById("demo").innerHTML = "TIME UP"; 
        document.getElementById("year").innerHTML ='0'; 
        document.getElementById("day").innerHTML ='0'; 
        document.getElementById("hour").innerHTML ='0'; 
        document.getElementById("minute").innerHTML ='0' ;  
        document.getElementById("second").innerHTML = '0'; } 
}, 1000);

有什么想法可以纠正这个问题并添加额外的天数来说明已经过去的闰年数吗？

非常感谢。

标签： javascriptdateleap-year

Answer 1

答案比我之前想象的要简单得多。

由于我只是想考虑闰年的天数差异，我只需要将年数除以 4 并加 1，如下所示：

var lastDay = new Date("Jan 1, 1994 10:00:00").getTime(); 
var x = setInterval(function() { 
var now = new Date().getTime(); 
var t = now - lastDay; 
var years = Math.floor(t / (1000 * 60 * 60 * 24)/ 365);
var leapDays = Math.floor((years / 4) + 1);
var days = Math.floor((t % (1000 * 60 * 60 * 24 * 365))/(1000 * 60 * 60 * 24) - leapDays); 
var hours = Math.floor((t % (1000 * 60 * 60 * 24))/(1000 * 60 * 60)); 
var minutes = Math.floor((t % (1000 * 60 * 60)) / (1000 * 60)); 
var seconds = Math.floor((t % (1000 * 60)) / 1000); 
document.getElementById("year").innerHTML =years ; 
document.getElementById("day").innerHTML =days ; 
document.getElementById("hour").innerHTML =hours; 
document.getElementById("minute").innerHTML = minutes;  
document.getElementById("second").innerHTML =seconds;  
if (t < 0) { 
        clearInterval(x); 
        document.getElementById("demo").innerHTML = "TIME UP"; 
        document.getElementById("year").innerHTML ='0'; 
        document.getElementById("day").innerHTML ='0'; 
        document.getElementById("hour").innerHTML ='0'; 
        document.getElementById("minute").innerHTML ='0' ;  
        document.getElementById("second").innerHTML = '0'; } 
}, 1000);