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

如果你想有一个DateArray日期试试这个:

<script>
        function getDates(startDate, stopDate) {
        var dateArray = new Array();
        var currentDate = moment(startDate);
        dateArray.push( moment(currentDate).format('L'));

        var stopDate = moment(stopDate);
        while (dateArray[dateArray.length -1] != stopDate._i) {
            dateArray.push( moment(currentDate).format('L'));
            currentDate = moment(currentDate).add(1, 'days');
        }
        return dateArray;
      }
</script>

调试片段

其他回答

Bookmarklet版本的其他答案,提示你两个日期:

javascript:(function() {
    var d = new Date(prompt("First Date or leave blank for today?") || Date.now());
    prompt("Days Between", Math.round(
        Math.abs(
            (d.getTime() - new Date(prompt("Date 2")).getTime())
                /(24*60*60*1000)
             )
        ));
})();

如果你有两个unix时间戳,你可以使用这个函数(为了清晰起见,写得更详细一点):

// Calculate number of days between two unix timestamps
// ------------------------------------------------------------
var daysBetween = function(timeStampA, timeStampB) {
    var oneDay = 24 * 60 * 60 * 1000; // hours * minutes * seconds * milliseconds
    var firstDate = new Date(timeStampA * 1000);
    var secondDate = new Date(timeStampB * 1000);
    var diffDays = Math.round(Math.abs((firstDate.getTime() - secondDate.getTime())/(oneDay)));
    return diffDays;
};

例子:

daysBetween(1096580303, 1308713220); // 2455

我也有同样的问题,但如果你在SQL查询上做的话会更好:

DateDiff(DAY, StartValue,GETDATE()) AS CountDays

查询将自动生成一个列CountDays

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;
    }
}

function timeDifference(date1, date2) { var oneDay = 24 * 60 * 60; // hours*minutes*seconds var oneHour = 60 * 60; // minutes*seconds var oneMinute = 60; // 60 seconds var firstDate = date1.getTime(); // convert to milliseconds var secondDate = date2.getTime(); // convert to milliseconds var seconds = Math.round(Math.abs(firstDate - secondDate) / 1000); //calculate the diffrence in seconds // the difference object var difference = { "days": 0, "hours": 0, "minutes": 0, "seconds": 0, } //calculate all the days and substract it from the total while (seconds >= oneDay) { difference.days++; seconds -= oneDay; } //calculate all the remaining hours then substract it from the total while (seconds >= oneHour) { difference.hours++; seconds -= oneHour; } //calculate all the remaining minutes then substract it from the total while (seconds >= oneMinute) { difference.minutes++; seconds -= oneMinute; } //the remaining seconds : difference.seconds = seconds; //return the difference object return difference; } console.log(timeDifference(new Date(2017,0,1,0,0,0),new Date()));