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
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">
有很多方法可以做到这一点。 是的，你可以使用普通的旧JS。试试:
let dt1 = new Date()
let dt2 = new Date()
让我们使用Date.prototype.setMinutes模拟通道，并确保我们在范围内。
dt1.setMinutes(7)
dt2.setMinutes(42)
console.log('Elapsed seconds:',(dt2-dt1)/1000)
或者你也可以使用一些像js-joda这样的库，在那里你可以很容易地做这样的事情(直接从文档中):
var dt1 = LocalDateTime.parse("2016-02-26T23:55:42.123");
var dt2 = dt1
.plusYears(6)
.plusMonths(12)
.plusHours(2)
.plusMinutes(42)
.plusSeconds(12);
// obtain the duration between the two dates
dt1.until(dt2, ChronoUnit.YEARS); // 7
dt1.until(dt2, ChronoUnit.MONTHS); // 84
dt1.until(dt2, ChronoUnit.WEEKS); // 356
dt1.until(dt2, ChronoUnit.DAYS); // 2557
dt1.until(dt2, ChronoUnit.HOURS); // 61370
dt1.until(dt2, ChronoUnit.MINUTES); // 3682242
dt1.until(dt2, ChronoUnit.SECONDS); // 220934532
有更多的ofc库，但js-joda还有一个额外的好处，它也可以在Java中使用，在Java中已经进行了广泛的测试。所有这些测试都已迁移到js-joda，它也是不可变的。
var date1 = new Date("06/30/2019");
var date2 = new Date("07/30/2019");
// To calculate the time difference of two dates
var Difference_In_Time = date2.getTime() - date1.getTime();
// To calculate the no. of days between two dates
var Difference_In_Days = Difference_In_Time / (1000 * 3600 * 24);
//To display the final no. of days (result)
document.write("Total number of days between dates <br>"
+ date1 + "<br> and <br>"
+ date2 + " is: <br> "
+ Difference_In_Days);
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。
