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

查看Date.js

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

setDate()在月份边界上有问题,在上面的注释中已经注意到。一个简单的解决方法是使用epoch时间戳来查找日期差异,而不是使用date对象上的方法(令人惊讶地违反直觉)。即。

function getPreviousMonday(fromDate) {
    var dayMillisecs = 24 * 60 * 60 * 1000;

    // Get Date object truncated to date.
    var d = new Date(new Date(fromDate || Date()).toISOString().slice(0, 10));

    // If today is Sunday (day 0) subtract an extra 7 days.
    var dayDiff = d.getDay() === 0 ? 7 : 0;

    // Get date diff in millisecs to avoid setDate() bugs with month boundaries.
    var mondayMillisecs = d.getTime() - (d.getDay() + dayDiff) * dayMillisecs;

    // Return date as YYYY-MM-DD string.
    return new Date(mondayMillisecs).toISOString().slice(0, 10);
}

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

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。