JavaScript - 从当前日期开始获得每周的第一天

Question

我需要最快的方法来获得一周的第一天。例如，今天是11月11日和星期四，我需要将该日期转换为11月8日（星期一）。我需要MongoDB地图功能最快的方法，有什么想法？

Answer 1

使用Date对象的getDay方法，你就可以知道星期几的数量（即0 =星期日，1 =星期一，等等）。

然后，您可以减去的天数加一，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 

Answer 2

退房Date.js

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

Answer 3

不知道它是如何比较的性能，但这个工程。

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 

或者作为一个函数：

function getMonday(date) { 
    var day = date.getDay() || 7; 
    if(day !== 1) 
     date.setHours(-24 * (day - 1)); 
    return date; 
} 

getMonday(new Date()); 

Answer 4

退房：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) 

奖励：可与node.js的太

Answer 5

我使用这个

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; 
} 

Answer 6

此功能使用当前毫秒时间减去当前周，再减去一个星期，如果当前日期是在周一（周日JavaScript的计数）。

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; 
} 

当周从一个月跨越到另一个（也是年）我已经测试过它，它似乎正常工作。

Answer 7

setDate（）对上面注释中提到的月边界有问题。一个干净的解决方法是使用纪元时间戳找到日期差异，而不是日期对象上的（令人惊讶的违反直觉的）方法。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); 
} 

Answer 8

晚上好，

我宁愿只是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)); 
} 

你会发现这其实利用了我用另一种扩展方法：

Date.prototype.addDays = function (pDays) { 
    var mDate = new Date(this.valueOf()); 
    mDate.setDate(mDate.getDate() + pDays); 
    return mDate; 
}; 

现在，如果您的星期从星期天开始，请为pStartOfWeek参数传递一个“0”，如下所示：

var mThisSunday = new Date().startOfWeek(0);

同样，如果周的星期一开始，传递一个为pStartOfWeek参数 “1”：

var mThisMonday = new Date().startOfWeek(1);

问候，

Answer 9

这里是我的解决方案：

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; 
} 

Now你可以通过返回的数组索引来选择日期。

Answer 10

var dt = new Date(); // current date of week 
var currentWeekDay = dt.getDay(); 
var lessDays = currentWeekDay == 0 ? 6 : currentWeekDay - 1; 
var wkStart = new Date(new Date(dt).setDate(dt.getDate() - lessDays)); 
var wkEnd = new Date(new Date(wkStart).setDate(wkStart.getDate() + 6)); 

这会很好用。