JavaScript Date.getWeek()? [duplicate]

I'm looking for a tested solid solution for getting current week of the year for specified date. All I can find are the ones that doesn't take in account leap years or just plain wrong. Does anyone have this type of stuff?

Or even better a function that says how many weeks does month occupy. It is usually 5, but can be 4 (feb) or 6 (1st is sunday and month has 30-31 days in it)

================= UPDATE:

Still not sure about getting week #, but since I figured out it won't solve my problem with calculating how many weeks month occupy, I abandoned it.

Here's a function to find out how many weeks exactly month occupy on the calendar:

getWeeksNum: function(year, month) {
    var daysNum = 32 - new Date(year, month, 32).getDate(),
        fDayO = new Date(year, month, 1).getDay(),
        fDay = fDayO ? (fDayO - 1) : 6,
        weeksNum = Math.ceil((daysNum + fDay) / 7);
    return weeksNum;
}

Answers:

Answer
/**
 * Returns the week number for this date.  dowOffset is the day of week the week
 * "starts" on for your locale - it can be from 0 to 6. If dowOffset is 1 (Monday),
 * the week returned is the ISO 8601 week number.
 * @param int dowOffset
 * @return int
 */
Date.prototype.getWeek = function (dowOffset) {
/*getWeek() was developed by Nick Baicoianu at MeanFreePath: http://www.meanfreepath.com */

    dowOffset = typeof(dowOffset) == 'int' ? dowOffset : 0; //default dowOffset to zero
    var newYear = new Date(this.getFullYear(),0,1);
    var day = newYear.getDay() - dowOffset; //the day of week the year begins on
    day = (day >= 0 ? day : day + 7);
    var daynum = Math.floor((this.getTime() - newYear.getTime() - 
    (this.getTimezoneOffset()-newYear.getTimezoneOffset())*60000)/86400000) + 1;
    var weeknum;
    //if the year starts before the middle of a week
    if(day < 4) {
        weeknum = Math.floor((daynum+day-1)/7) + 1;
        if(weeknum > 52) {
            nYear = new Date(this.getFullYear() + 1,0,1);
            nday = nYear.getDay() - dowOffset;
            nday = nday >= 0 ? nday : nday + 7;
            /*if the next year starts before the middle of
              the week, it is week #1 of that year*/
            weeknum = nday < 4 ? 1 : 53;
        }
    }
    else {
        weeknum = Math.floor((daynum+day-1)/7);
    }
    return weeknum;
};

Usage:

var mydate = new Date(2011,2,3); // month number starts from 0
// or like this
var mydate = new Date('March 3, 2011');
alert(mydate.getWeek());

Source

Answer

Get week number

Date.prototype.getWeek = function() {
    var dt = new Date(this.getFullYear(),0,1);
    return Math.ceil((((this - dt) / 86400000) + dt.getDay()+1)/7);
};

var myDate = new Date(2013, 3, 25); // 2013, 25 April

console.log(myDate.getWeek());
Answer

For those looking for a more simple approach;

Date.prototype.getWeek = function() {
  var onejan = new Date(this.getFullYear(),0,1);
  var today = new Date(this.getFullYear(),this.getMonth(),this.getDate());
  var dayOfYear = ((today - onejan + 86400000)/86400000);
  return Math.ceil(dayOfYear/7)
};

Use with:

var today = new Date();
var currentWeekNumber = today.getWeek();
console.log(currentWeekNumber);
Answer

Consider using my implementation of "Date.prototype.getWeek", think is more accurate than the others i have seen here :)

Date.prototype.getWeek = function(){
    // We have to compare against the first monday of the year not the 01/01
    // 60*60*24*1000 = 86400000
    // 'onejan_next_monday_time' reffers to the miliseconds of the next monday after 01/01

    var day_miliseconds = 86400000,
        onejan = new Date(this.getFullYear(),0,1,0,0,0),
        onejan_day = (onejan.getDay()==0) ? 7 : onejan.getDay(),
        days_for_next_monday = (8-onejan_day),
        onejan_next_monday_time = onejan.getTime() + (days_for_next_monday * day_miliseconds),
        // If one jan is not a monday, get the first monday of the year
        first_monday_year_time = (onejan_day>1) ? onejan_next_monday_time : onejan.getTime(),
        this_date = new Date(this.getFullYear(), this.getMonth(),this.getDate(),0,0,0),// This at 00:00:00
        this_time = this_date.getTime(),
        days_from_first_monday = Math.round(((this_time - first_monday_year_time) / day_miliseconds));

    var first_monday_year = new Date(first_monday_year_time);

    // We add 1 to "days_from_first_monday" because if "days_from_first_monday" is *7,
    // then 7/7 = 1, and as we are 7 days from first monday,
    // we should be in week number 2 instead of week number 1 (7/7=1)
    // We consider week number as 52 when "days_from_first_monday" is lower than 0,
    // that means the actual week started before the first monday so that means we are on the firsts
    // days of the year (ex: we are on Friday 01/01, then "days_from_first_monday"=-3,
    // so friday 01/01 is part of week number 52 from past year)
    // "days_from_first_monday<=364" because (364+1)/7 == 52, if we are on day 365, then (365+1)/7 >= 52 (Math.ceil(366/7)=53) and thats wrong

    return (days_from_first_monday>=0 && days_from_first_monday<364) ? Math.ceil((days_from_first_monday+1)/7) : 52;
}

You can check my public repo here https://bitbucket.org/agustinhaller/date.getweek (Tests included)

Answer

I know this is an old question, but maybe it helps:

http://weeknumber.net/how-to/javascript

Answer

/*get the week number by following the norms of ISO 8601*/
function getWeek(dt){
	var calc=function(o){
		if(o.dtmin.getDay()!=1){
			if(o.dtmin.getDay()<=4 && o.dtmin.getDay()!=0)o.w+=1;
			o.dtmin.setDate((o.dtmin.getDay()==0)? 2 : 1+(7-o.dtmin.getDay())+1);
		}
		o.w+=Math.ceil((((o.dtmax.getTime()-o.dtmin.getTime())/(24*60*60*1000))+1)/7);
	},getNbDaysInAMonth=function(year,month){
		var nbdays=31;
		for(var i=0;i<=3;i++){
			nbdays=nbdays-i;
			if((dtInst=new Date(year,month-1,nbdays)) && dtInst.getDate()==nbdays && (dtInst.getMonth()+1)==month  && dtInst.getFullYear()==year)
				break;
		}
		return nbdays;
	};
	if(dt.getMonth()+1==1 && dt.getDate()>=1 && dt.getDate()<=3 && (dt.getDay()>=5 || dt.getDay()==0)){
		var pyData={"dtmin":new Date(dt.getFullYear()-1,0,1,0,0,0,0),"dtmax":new Date(dt.getFullYear()-1,11,getNbDaysInAMonth(dt.getFullYear()-1,12),0,0,0,0),"w":0};
		calc(pyData);
		return pyData.w;
	}else{
		var ayData={"dtmin":new Date(dt.getFullYear(),0,1,0,0,0,0),"dtmax":new Date(dt.getFullYear(),dt.getMonth(),dt.getDate(),0,0,0,0),"w":0},
			nd12m=getNbDaysInAMonth(dt.getFullYear(),12);
		if(dt.getMonth()==12 && dt.getDay()!=0 && dt.getDay()<=3 && nd12m-dt.getDate()<=3-dt.getDay())ayData.w=1;else calc(ayData);
		return ayData.w;
	}
}
alert(getWeek(new Date(2017,01-1,01)));

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