var start= $("#firstDate").datepicker("getDate");
var end= $("#SecondDate").datepicker("getDate");
var days = (end- start) / (1000 * 60 * 60 * 24);
 alert(Math.round(days));

Jsfiddle示例:)

 // JavaScript / NodeJs answer  
   let startDate = new Date("2022-09-19");
   let endDate = new Date("2022-09-26");

   let difference = startDate.getTime() - endDate.getTime();
   
    console.log(difference);

   let TotalDiffDays = Math.ceil(difference / (1000 * 3600 * 24));
   console.log(TotalDiffDays + " days :) ");

我认为解决方案不是100%正确的，我会使用天花板而不是地板，圆形将工作，但这不是正确的操作。

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

function formatDate(seconds, dictionary) {
    var foo = new Date;
    var unixtime_ms = foo.getTime();
    var unixtime = parseInt(unixtime_ms / 1000);
    var diff = unixtime - seconds;
    var display_date;
    if (diff <= 0) {
        display_date = dictionary.now;
    } else if (diff < 60) {
        if (diff == 1) {
            display_date = diff + ' ' + dictionary.second;
        } else {
            display_date = diff + ' ' + dictionary.seconds;
        }
    } else if (diff < 3540) {
        diff = Math.round(diff / 60);
        if (diff == 1) {
            display_date = diff + ' ' + dictionary.minute;
        } else {
            display_date = diff + ' ' + dictionary.minutes;
        }
    } else if (diff < 82800) {
        diff = Math.round(diff / 3600);
        if (diff == 1) {
            display_date = diff + ' ' + dictionary.hour;
        } else {
            display_date = diff + ' ' + dictionary.hours;
        }
    } else {
        diff = Math.round(diff / 86400);
        if (diff == 1) {
            display_date = diff + ' ' + dictionary.day;
        } else {
            display_date = diff + ' ' + dictionary.days;
        }
    }
    return display_date;
}

我从其他答案中得到一些灵感，使输入具有自动卫生。我希望这是对其他答案的改进。

我还推荐使用<input type="date">字段，这将有助于验证用户输入。

//use best practices by labeling your constants. let MS_PER_SEC = 1000 , SEC_PER_HR = 60 * 60 , HR_PER_DAY = 24 , MS_PER_DAY = MS_PER_SEC * SEC_PER_HR * HR_PER_DAY ; //let's assume we get Date objects as arguments, otherwise return 0. function dateDiffInDays(date1Time, date2Time) { if (!date1Time || !date2Time) return 0; return Math.round((date2Time - date1Time) / MS_PER_DAY); } function getUTCTime(dateStr) { const date = new Date(dateStr); // If use 'Date.getTime()' it doesn't compute the right amount of days // if there is a 'day saving time' change between dates return Date.UTC(date.getFullYear(), date.getMonth(), date.getDate()); } function calcInputs() { let date1 = document.getElementById("date1") , date2 = document.getElementById("date2") , resultSpan = document.getElementById("result") ; if (date1.value && date2.value && resultSpan) { //remove non-date characters console.log(getUTCTime(date1.value)); let date1Time = getUTCTime(date1.value) , date2Time = getUTCTime(date2.value) , result = dateDiffInDays(date1Time, date2Time) ; resultSpan.innerHTML = result + " days"; } } window.onload = function() { calcInputs(); }; //some code examples console.log(dateDiffInDays(new Date("1/15/2019"), new Date("1/30/2019"))); console.log(dateDiffInDays(new Date("1/15/2019"), new Date("2/30/2019"))); console.log(dateDiffInDays(new Date("1/15/2000"), new Date("1/15/2019"))); <input type="date" id="date1" value="2000-01-01" onchange="calcInputs();" /> <input type="date" id="date2" value="2022-01-01" onchange="calcInputs();"/> Result: <span id="result"></span>

当我想在两个日期上做一些计算时，我发现了这个问题，但是日期有小时和分钟的值，我修改了@michael-liu的答案来满足我的要求，它通过了我的测试。

差异日期2012-12-31 23:00和2013-01-01 01:00应该等于1。(2小时) 差异日期2012-12-31 01:00和2013-01-01 23:00应该等于1。(46个小时)

function treatAsUTC(date) {
    var result = new Date(date);
    result.setMinutes(result.getMinutes() - result.getTimezoneOffset());
    return result;
}

var millisecondsPerDay = 24 * 60 * 60 * 1000;
function diffDays(startDate, endDate) {
    return Math.floor(treatAsUTC(endDate) / millisecondsPerDay) - Math.floor(treatAsUTC(startDate) / millisecondsPerDay);
}