扩展回答来自@Christian C. Salvadó和来自@Ayyash(对象是可变的)和@Awi和@Louis Ameline(设置时间为00:00:00)的信息

函数可以是这样的

function getMonday(d) {
  var day = d.getDay(),
      diff = d.getDate() - day + (day == 0 ? -6:1); // adjust when day is sunday
  d.setDate(diff);
  d.setHours(0,0,0,0); // set hours to 00:00:00

  return d; // object is mutable no need to recreate object
}

getMonday(new Date())

setDate()在月份边界上有问题，在上面的注释中已经注意到。一个简单的解决方法是使用epoch时间戳来查找日期差异，而不是使用date对象上的方法(令人惊讶地违反直觉)。即。

function getPreviousMonday(fromDate) {
    var dayMillisecs = 24 * 60 * 60 * 1000;

    // Get Date object truncated to date.
    var d = new Date(new Date(fromDate || Date()).toISOString().slice(0, 10));

    // If today is Sunday (day 0) subtract an extra 7 days.
    var dayDiff = d.getDay() === 0 ? 7 : 0;

    // Get date diff in millisecs to avoid setDate() bugs with month boundaries.
    var mondayMillisecs = d.getTime() - (d.getDay() + dayDiff) * dayMillisecs;

    // Return date as YYYY-MM-DD string.
    return new Date(mondayMillisecs).toISOString().slice(0, 10);
}

CMS的答案是正确的，但假设星期一是一周的第一天。 钱德勒·兹沃勒的答案是正确的，但摆弄了日期原型。 其他加/减小时/分钟/秒/毫秒的答案是错误的，因为不是所有的日子都有24小时。

下面的函数是正确的，它将日期作为第一个参数，将所需的一周第一天作为第二个参数(0表示周日，1表示周一，等等)。注意:小时、分、秒设置为0才有一天的开始。

function firstDayOfWeek(dateObject, firstDayOfWeekIndex) { const dayOfWeek = dateObject.getDay(), firstDayOfWeek = new Date(dateObject), diff = dayOfWeek >= firstDayOfWeekIndex ? dayOfWeek - firstDayOfWeekIndex : 6 - dayOfWeek firstDayOfWeek.setDate(dateObject.getDate() - diff) firstDayOfWeek.setHours(0,0,0,0) return firstDayOfWeek } // August 18th was a Saturday let lastMonday = firstDayOfWeek(new Date('August 18, 2018 03:24:00'), 1) // outputs something like "Mon Aug 13 2018 00:00:00 GMT+0200" // (may vary according to your time zone) document.write(lastMonday)

周一上午00点到周一上午00点返回。

const now = new Date()
const startOfWeek = new Date(now.getFullYear(), now.getMonth(), now.getDate() - now.getDay() + 1)
const endOfWeek = new Date(now.getFullYear(), now.getMonth(), startOfWeek.getDate() + 7)

查看Date.js

Date.today().previous().monday()

以下是我的解决方案:

function getWeekDates(){
    var day_milliseconds = 24*60*60*1000;
    var dates = [];
    var current_date = new Date();
    var monday = new Date(current_date.getTime()-(current_date.getDay()-1)*day_milliseconds);
    var sunday = new Date(monday.getTime()+6*day_milliseconds);
    dates.push(monday);
    for(var i = 1; i < 6; i++){
        dates.push(new Date(monday.getTime()+i*day_milliseconds));
    }
    dates.push(sunday);
    return dates;
}

现在你可以通过返回的数组索引来选择日期。