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)

晚上好,

我更喜欢有一个简单的扩展方法:

Date.prototype.startOfWeek = function (pStartOfWeek) {
    var mDifference = this.getDay() - pStartOfWeek;

    if (mDifference < 0) {
        mDifference += 7;
    }

    return new Date(this.addDays(mDifference * -1));
}

你会注意到这实际上利用了我使用的另一个扩展方法:

Date.prototype.addDays = function (pDays) {
    var mDate = new Date(this.valueOf());
    mDate.setDate(mDate.getDate() + pDays);
    return mDate;
};

现在，如果你的周从周日开始，为pStartOfWeek参数传递一个“0”，如下所示:

var mThisSunday = new Date（）.startOfWeek（0）;

类似地，如果你的周从星期一开始，为pStartOfWeek参数传递一个“1”:

var mThisMonday = new Date（）.startOfWeek（1）;

问候,

该函数使用当前毫秒时间减去当前周，如果当前日期是周一，则再减去一周(javascript从周日开始计数)。

function getMonday(fromDate) {
    // length of one day i milliseconds
  var dayLength = 24 * 60 * 60 * 1000;

  // Get the current date (without time)
    var currentDate = new Date(fromDate.getFullYear(), fromDate.getMonth(), fromDate.getDate());

  // Get the current date's millisecond for this week
  var currentWeekDayMillisecond = ((currentDate.getDay()) * dayLength);

  // subtract the current date with the current date's millisecond for this week
  var monday = new Date(currentDate.getTime() - currentWeekDayMillisecond + dayLength);

  if (monday > currentDate) {
    // It is sunday, so we need to go back further
    monday = new Date(monday.getTime() - (dayLength * 7));
  }

  return monday;
}

当一周从一个月延伸到另一个月(也包括几年)时，我对它进行了测试，它似乎可以正常工作。

查看:moment.js

例子:

moment().day(-7); // last Sunday (0 - 7)
moment().day(7); // next Sunday (0 + 7)
moment().day(10); // next Wednesday (3 + 7)
moment().day(24); // 3 Wednesdays from now (3 + 7 + 7 + 7)

好处:也适用于node.js

使用Date对象的getDay方法，您可以知道一周中的天数(0=星期日，1=星期一，等等)。

然后你可以用这个天数加1，例如:

function getMonday(d) {
  d = new Date(d);
  var day = d.getDay(),
      diff = d.getDate() - day + (day == 0 ? -6:1); // adjust when day is sunday
  return new Date(d.setDate(diff));
}

getMonday(new Date()); // Mon Nov 08 2010

接受的答案将不适用于在UTC-XX:XX时区运行代码的任何人。

这里的代码将工作，无论时区仅为日期。如果你也提供时间，这就行不通了。只提供日期或解析日期并将其作为输入。我在代码开始时提到了不同的测试用例。

function getDateForTheMonday(dateString) { var orignalDate = new Date(dateString) var modifiedDate = new Date(dateString) var day = modifiedDate.getDay() diff = modifiedDate.getDate() - day + (day == 0 ? -6:1);// adjust when day is sunday modifiedDate.setDate(diff) var diffInDate = orignalDate.getDate() - modifiedDate.getDate() if(diffInDate == 6) { diff = diff + 7 modifiedDate.setDate(diff) } console.log("Given Date : " + orignalDate.toUTCString()) console.log("Modified date for Monday : " + modifiedDate) } getDateForTheMonday("2022-08-01") // Jul month with 31 Days getDateForTheMonday("2022-07-01") // June month with 30 days getDateForTheMonday("2022-03-01") // Non leap year February getDateForTheMonday("2020-03-01") // Leap year February getDateForTheMonday("2022-01-01") // First day of the year getDateForTheMonday("2021-12-31") // Last day of the year