如何获取JavaScript中两个日期之间的秒数？

Question

我正在尝试获取JavaScript中两个日期之间的秒数。
这里和这里的问题建议一个简单的减法运算就可以解决问题，而其他答案则建议您使用.getTime()或.valueOf()属性。
但是尽管如此，当我减去两个相距5秒的日期时，我得到的值1551181475795太大而无法正确。
这是我的代码：

 var countDownDate = new Date("Mar 16, 2019 16:33:25").valueOf(); var dist = 347155200; // Update the count down every 1 second var x = setInterval(function() { // Get todays date and time var now = new Date("Mar 16, 2019 16:33:30"); // Find the distance between now and the count down date var distance = Math.abs(now - countDownDate / 1000); // Display the result in the element with id="demo" document.getElementById("demo").innerHTML = distance; // If the count down is finished, write some text if (distance < 0) { clearInterval(x); document.getElementById("demo").innerHTML = "EXPIRED"; } }, 1000); 

 <p id='demo'> 

如您所见，两个时间仅相隔5秒，但将它们分开的值没有意义。
提前致谢！

Answer 1

这是一个操作顺序，在进行减法之前，您要将countDownDate除以1000 （现在正在计算now - countDownDate / 1000但是应该先计算(now - countDownDate) / 1000 ，先减去）：

 var countDownDate = new Date("Mar 16, 2019 16:33:25").valueOf(); var dist = 347155200; // Update the count down every 1 second var x = setInterval(function() { // Get todays date and time var now = new Date("Mar 16, 2019 16:33:30"); // Find the distance between now and the count down date var distance = Math.abs((now - countDownDate) / 1000); // Display the result in the element with id="demo" document.getElementById("demo").innerHTML = distance; // If the count down is finished, write some text if (distance < 0) { clearInterval(x); document.getElementById("demo").innerHTML = "EXPIRED"; } }, 1000); 

 <p id="demo"></p> 

Answer 2

如您所研究的，您可以简单地将较旧的日期从较旧的日期中减少，该较旧的日期返回两个日期之间的毫秒数。

用1000除以得到秒数。

 let start = new Date("Mar 16, 2019 16:33:25") let end = new Date("Mar 16, 2019 16:33:30"); let secondsBetween = (end - start) / 1000; console.log("without value of:", secondsBetween, "seconds") start = new Date("Mar 16, 2019 16:33:25").valueOf() end = new Date("Mar 16, 2019 16:33:30").valueOf() secondsBetween = (end - start) / 1000; console.log("using value of:",(end - start) / 1000, "seconds")