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
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());
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。
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
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
我在用这个
function get_next_week_start() {
var now = new Date();
var next_week_start = new Date(now.getFullYear(), now.getMonth(), now.getDate()+(8 - now.getDay()));
return next_week_start;
}