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,

查看Date.js

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

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。

一周的第一天/最后一天

为了得到即将到来的一周的第一天，你可以这样使用:

函数getUpcomingSunday() { const date = new date (); const today = date.getDate(); const currentDay = date.getDay(); const newDate =日期。setDate(今天- currentDay + 7); return newDate (newDate); ｝ console.log (getUpcomingSunday ());

或者获得最晚的第一天:

函数getLastSunday() { const date = new date (); const today = date.getDate(); const currentDay = date.getDay(); const newDate =日期。setDate(今天- (currentDay || 7)); return newDate (newDate); ｝ console.log (getLastSunday ());

*根据您所在的时区，一周的开始不必从周日开始，它可以从周五、周六、周一或您的机器设置的任何其他日子开始。这些方法可以解释。

*你也可以像这样使用toISOString方法格式化它:

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

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())