我需要最快的方法得到一周的第一天。例如:今天是11月11日,是星期四;我想要这周的第一天,也就是11月8日,一个星期一。我需要MongoDB映射函数的最快方法,有什么想法吗?


当前回答

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

其他回答

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。

我用这个:

let current_date = new Date();
let days_to_monday = 1 - current_date.getDay();
monday_date = current_date.addDays(days_to_monday);

// https://stackoverflow.com/a/563442/6533037
Date.prototype.addDays = function(days) {
    var date = new Date(this.valueOf());
    date.setDate(date.getDate() + days);
    return date;
}

它工作得很好。

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())

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,

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。