我想用天、小时、分钟、秒、毫秒、纳秒来计算日期差异。我该怎么做呢?
当前回答
function daysInMonth (month, year) { return new Date(year, month, 0).getDate(); } function getduration(){ let A= document.getElementById("date1_id").value let B= document.getElementById("date2_id").value let C=Number(A.substring(3,5)) let D=Number(B.substring(3,5)) let dif=D-C let arr=[]; let sum=0; for (let i=0;i<dif+1;i++){ sum+=Number(daysInMonth(i+C,2019)) } let sum_alter=0; for (let i=0;i<dif;i++){ sum_alter+=Number(daysInMonth(i+C,2019)) } let no_of_month=(Number(B.substring(3,5)) - Number(A.substring(3,5))) let days=[]; if ((Number(B.substring(3,5)) - Number(A.substring(3,5)))>0||Number(B.substring(0,2)) - Number(A.substring(0,2))<0){ days=Number(B.substring(0,2)) - Number(A.substring(0,2)) + sum_alter } if ((Number(B.substring(3,5)) == Number(A.substring(3,5)))){ console.log(Number(B.substring(0,2)) - Number(A.substring(0,2)) + sum_alter) } time_1=[]; time_2=[]; let hour=[]; time_1=document.getElementById("time1_id").value time_2=document.getElementById("time2_id").value if (time_1.substring(0,2)=="12"){ time_1="00:00:00 PM" } if (time_1.substring(9,11)==time_2.substring(9,11)){ hour=Math.abs(Number(time_2.substring(0,2)) - Number(time_1.substring(0,2))) } if (time_1.substring(9,11)!=time_2.substring(9,11)){ hour=Math.abs(Number(time_2.substring(0,2)) - Number(time_1.substring(0,2)))+12 } let min=Math.abs(Number(time_1.substring(3,5))-Number(time_2.substring(3,5))) document.getElementById("duration_id").value=days +" days "+ hour+" hour " + min+" min " } <input type="text" id="date1_id" placeholder="28/05/2019"> <input type="text" id="date2_id" placeholder="29/06/2019"> <br><br> <input type="text" id="time1_id" placeholder="08:01:00 AM"> <input type="text" id="time2_id" placeholder="00:00:00 PM"> <br><br> <button class="text" onClick="getduration()">Submit </button> <br><br> <input type="text" id="duration_id" placeholder="days hour min">
其他回答
<html lang="en">
<head>
<script>
function getDateDiff(time1, time2) {
var str1= time1.split('/');
var str2= time2.split('/');
// yyyy , mm , dd
var t1 = new Date(str1[2], str1[0]-1, str1[1]);
var t2 = new Date(str2[2], str2[0]-1, str2[1]);
var diffMS = t1 - t2;
console.log(diffMS + ' ms');
var diffS = diffMS / 1000;
console.log(diffS + ' ');
var diffM = diffS / 60;
console.log(diffM + ' minutes');
var diffH = diffM / 60;
console.log(diffH + ' hours');
var diffD = diffH / 24;
console.log(diffD + ' days');
alert(diffD);
}
//alert(getDateDiff('10/18/2013','10/14/2013'));
</script>
</head>
<body>
<input type="button"
onclick="getDateDiff('10/18/2013','10/14/2013')"
value="clickHere()" />
</body>
</html>
这就是如何在没有框架的情况下实现日期之间的差异。
function getDateDiff(dateOne, dateTwo) {
if(dateOne.charAt(2)=='-' & dateTwo.charAt(2)=='-'){
dateOne = new Date(formatDate(dateOne));
dateTwo = new Date(formatDate(dateTwo));
}
else{
dateOne = new Date(dateOne);
dateTwo = new Date(dateTwo);
}
let timeDiff = Math.abs(dateOne.getTime() - dateTwo.getTime());
let diffDays = Math.ceil(timeDiff / (1000 * 3600 * 24));
let diffMonths = Math.ceil(diffDays/31);
let diffYears = Math.ceil(diffMonths/12);
let message = "Difference in Days: " + diffDays + " " +
"Difference in Months: " + diffMonths+ " " +
"Difference in Years: " + diffYears;
return message;
}
function formatDate(date) {
return date.split('-').reverse().join('-');
}
console.log(getDateDiff("23-04-2017", "23-04-2018"));
假设你有两个Date对象，你可以减去它们，以毫秒为单位得到差值:
var difference = date2 - date1;
从那里，您可以使用简单的算术来推导其他值。
另一种解决方案是将difference转换为一个新的Date对象，并获得该日期的年(与1970年不同)、月、日等。
var date1 = new Date(2010, 6, 17);
var date2 = new Date(2013, 12, 18);
var diff = new Date(date2.getTime() - date1.getTime());
// diff is: Thu Jul 05 1973 04:00:00 GMT+0300 (EEST)
console.log(diff.getUTCFullYear() - 1970); // Gives difference as year
// 3
console.log(diff.getUTCMonth()); // Gives month count of difference
// 6
console.log(diff.getUTCDate() - 1); // Gives day count of difference
// 4
所以差异就像“3年6个月零4天”。如果你想以一种人类可读的风格呈现不同，这将对你有所帮助。
使用Moment.js进行所有与JavaScript相关的日期时间计算
你问题的答案是:
var a = moment([2007, 0, 29]);
var b = moment([2007, 0, 28]);
a.diff(b) // 86400000
完整的细节可以在这里找到
