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
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,
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
);
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。
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)
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