例如,在输入框中给定两个日期:

<input id="first" value="1/1/2000"/>
<input id="second" value="1/1/2001"/>

<script>
  alert(datediff("day", first, second)); // what goes here?
</script>

如何在JavaScript中获得两个日期之间的天数?


当前回答

这可能不是最优雅的解决方案,但我认为它似乎用一段相对简单的代码就回答了这个问题。你不能用这样的词吗?

function dayDiff(startdate, enddate) {
  var dayCount = 0;

  while(enddate >= startdate) {
    dayCount++;
    startdate.setDate(startdate.getDate() + 1);
  }

return dayCount; 
}

这是假设您将日期对象作为参数传递。

其他回答

区分两个日期的最简单方法是:

var diff = Math.floor((Date.parse(str2) - Date.parse(str1)) / 86400000);

您将得到不同的天数(如果其中一个或两个都无法解析,则为NaN)。解析日期给出了以毫秒为单位的结果,要按天得到它,你必须除以24 * 60 * 60 * 1000

如果你想用天、小时、分钟、秒和毫秒来划分:

function dateDiff( str1, str2 ) {
    var diff = Date.parse( str2 ) - Date.parse( str1 ); 
    return isNaN( diff ) ? NaN : {
        diff : diff,
        ms : Math.floor( diff            % 1000 ),
        s  : Math.floor( diff /     1000 %   60 ),
        m  : Math.floor( diff /    60000 %   60 ),
        h  : Math.floor( diff /  3600000 %   24 ),
        d  : Math.floor( diff / 86400000        )
    };
}

以下是我对James版本的重构版本:

function mydiff(date1,date2,interval) {
    var second=1000, minute=second*60, hour=minute*60, day=hour*24, week=day*7;
    date1 = new Date(date1);
    date2 = new Date(date2);
    var timediff = date2 - date1;
    if (isNaN(timediff)) return NaN;
    switch (interval) {
        case "years": return date2.getFullYear() - date1.getFullYear();
        case "months": return (
            ( date2.getFullYear() * 12 + date2.getMonth() )
            -
            ( date1.getFullYear() * 12 + date1.getMonth() )
        );
        case "weeks"  : return Math.floor(timediff / week);
        case "days"   : return Math.floor(timediff / day); 
        case "hours"  : return Math.floor(timediff / hour); 
        case "minutes": return Math.floor(timediff / minute);
        case "seconds": return Math.floor(timediff / second);
        default: return undefined;
    }
}

最好还是取消夏令时吧,马斯。装天花板,数学。楼层等,使用UTC时间:

var firstDate = Date.UTC(2015,01,2);
var secondDate = Date.UTC(2015,04,22);
var diff = Math.abs((firstDate.valueOf() 
    - secondDate.valueOf())/(24*60*60*1000));

这个例子给出了109天的差异。24*60*60*1000是一天,单位是毫秒。

在这种情况下使用moment会容易得多,你可以试试这个:

    let days = moment(yourFirstDateString).diff(moment(yourSecondDateString), 'days');

它会给你一个整数值,比如1、2、5、0等,所以你可以很容易地使用条件检查,比如:

if(days < 1) {

此外,还有一件事是你可以得到更准确的时间差结果(以小数形式,如1.2,1.5,0.7等),使用以下语法得到这种结果:

let days = moment(yourFirstDateString).diff(moment(yourSecondDateString), 'days', true);

如果你有任何进一步的疑问,请告诉我

   function validateDate() {
        // get dates from input fields
        var startDate = $("#startDate").val();
        var endDate = $("#endDate").val();
        var sdate = startDate.split("-");
        var edate = endDate.split("-");
        var diffd = (edate[2] - sdate[2]) + 1;
        var leap = [ 0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 ];
        var nonleap = [ 0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 ];
        if (sdate[0] > edate[0]) {
            alert("Please enter End Date Year greater than Start Date Year");
            document.getElementById("endDate").value = "";
            diffd = "";
        } else if (sdate[1] > edate[1]) {
            alert("Please enter End Date month greater than Start Date month");
            document.getElementById("endDate").value = "";
            diffd = "";
        } else if (sdate[2] > edate[2]) {
            alert("Please enter End Date greater than Start Date");
            document.getElementById("endDate").value = "";
            diffd = "";
        } else {
            if (sdate[0] / 4 == 0) {
                while (sdate[1] < edate[1]) {
                    diffd = diffd + leap[sdate[1]++];
                }
            } else {
                while (sdate[1] < edate[1]) {
                    diffd = diffd + nonleap[sdate[1]++];
                }
            }
            document.getElementById("numberOfDays").value = diffd;
        }
    }

我建议使用moment.js库(http://momentjs.com/docs/#/displaying/difference/)。它正确地处理夏令时,通常是很好的工作。

例子:

var start = moment("2013-11-03");
var end = moment("2013-11-04");
end.diff(start, "days")
1