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

区分本地时间和UTC时间是很重要的。我想用UTC找到一周的开始,所以我使用了下面的函数。

function start_of_week_utc(date, start_day = 1) {

// Returns the start of the week containing a 'date'. Monday 00:00 UTC is
// considered to be the boundary between adjacent weeks, unless 'start_day' is
// specified. A Date object is returned.

    date = new Date(date);
    const day_of_month = date.getUTCDate();
    const day_of_week = date.getUTCDay();
    const difference_in_days = (
        day_of_week >= start_day
        ? day_of_week - start_day
        : day_of_week - start_day + 7
    );
    date.setUTCDate(day_of_month - difference_in_days);
    date.setUTCHours(0);
    date.setUTCMinutes(0);
    date.setUTCSeconds(0);
    date.setUTCMilliseconds(0);
    return date;
}

要在给定时区中找到一周的开始,首先将时区偏移量添加到输入日期,然后从输出日期中减去时区偏移量。

const local_start_of_week = new Date(
    start_of_week_utc(
        date.getTime() + timezone_offset_ms
    ).getTime() - timezone_offset_ms
);

其他回答

区分本地时间和UTC时间是很重要的。我想用UTC找到一周的开始,所以我使用了下面的函数。

function start_of_week_utc(date, start_day = 1) {

// Returns the start of the week containing a 'date'. Monday 00:00 UTC is
// considered to be the boundary between adjacent weeks, unless 'start_day' is
// specified. A Date object is returned.

    date = new Date(date);
    const day_of_month = date.getUTCDate();
    const day_of_week = date.getUTCDay();
    const difference_in_days = (
        day_of_week >= start_day
        ? day_of_week - start_day
        : day_of_week - start_day + 7
    );
    date.setUTCDate(day_of_month - difference_in_days);
    date.setUTCHours(0);
    date.setUTCMinutes(0);
    date.setUTCSeconds(0);
    date.setUTCMilliseconds(0);
    return date;
}

要在给定时区中找到一周的开始,首先将时区偏移量添加到输入日期,然后从输出日期中减去时区偏移量。

const local_start_of_week = new Date(
    start_of_week_utc(
        date.getTime() + timezone_offset_ms
    ).getTime() - timezone_offset_ms
);

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)

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())

一个只有数学计算的例子,没有任何Date函数。

const date = new date (); Const ts = +日期; const mondayTS = ts % (60 * 60 * 24 * (7-4) * 1000); const monday =新的日期(星期一); console.log(monday.toISOString(), 'Day:', monday.getDay());

const formatTS = v => new Date(v).toISOString(); const adjust = (v, d = 1) => v - v % (d * 1000); const d = new Date('2020-04-22T21:48:17.468Z'); const ts = +d; // 1587592097468 const test = v => console.log(formatTS(adjust(ts, v))); test(); // 2020-04-22T21:48:17.000Z test(60); // 2020-04-22T21:48:00.000Z test(60 * 60); // 2020-04-22T21:00:00.000Z test(60 * 60 * 24); // 2020-04-22T00:00:00.000Z test(60 * 60 * 24 * (7-4)); // 2020-04-20T00:00:00.000Z, monday // So, what does `(7-4)` mean? // 7 - days number in the week // 4 - shifting for the weekday number of the first second of the 1970 year, the first time stamp second. // new Date(0) ---> 1970-01-01T00:00:00.000Z // new Date(0).getDay() ---> 4

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,