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)

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

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

更普遍的说法是……这将根据您指定的日期给出当前一周中的任何一天。

//返回一周中的相对日期0 = Sunday, 1 = Monday…6 =星期六 函数getRelativeDayInWeek(d,dy) { d = new日期(d); var day = d.getDay()， diff = d.getDate() - day + (day == 0 ?6: dy);//当白天是星期天时进行调整 (d.setDate(diff)); ｝ var monday = getRelativeDayInWeek(new Date()，1); var friday = getRelativeDayInWeek(new Date()，5); console.log(星期一); console.log(周五);

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

不知道它的性能如何，但这是可行的。

var today = new Date();
var day = today.getDay() || 7; // Get current day number, converting Sun. to 7
if( day !== 1 )                // Only manipulate the date if it isn't Mon.
    today.setHours(-24 * (day - 1));   // Set the hours to day number minus 1
                                         //   multiplied by negative 24
alert(today); // will be Monday

或作为一个函数:

# modifies _date_
function setToMonday( date ) {
    var day = date.getDay() || 7;  
    if( day !== 1 ) 
        date.setHours(-24 * (day - 1)); 
    return date;
}

setToMonday(new Date());