JavaScript Date.getWeek()？
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JavaScript Date.getWeek()?
提问by Max
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?
我正在寻找一个经过测试的可靠解决方案，用于获取指定日期的当前周。我能找到的只是那些没有考虑闰年或完全错误的。有人有这种类型的东西吗？
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)
或者甚至更好的函数说明月份占用了多少周。它通常是 5，但可以是 4（二月）或 6（第 1 天是星期日，一个月有 30-31 天）
================= UPDATE:
================== 更新：
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.
仍然不确定获得第 # 周，但由于我发现它无法解决我计算一个月占用多少周的问题，因此我放弃了它。
Here's a function to find out how many weeks exactly month occupy on the calendar:
这是一个函数，用于找出日历上一个月究竟占据了多少周：
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;
}
回答by Cheery
/**
* 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:
用法：
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());
回答by alias51
For those looking for a more simple approach;
对于那些寻求更简单方法的人；
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:
与：
var today = new Date();
var currentWeekNumber = today.getWeek();
console.log(currentWeekNumber);
回答by Agustin Haller
Consider using my implementation of "Date.prototype.getWeek", think is more accurate than the others i have seen here :)
考虑使用我的“Date.prototype.getWeek”实现，认为比我在这里看到的其他人更准确:)
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)
你可以在这里查看我的公共仓库https://bitbucket.org/agustinhaller/date.getweek（包括测试）
回答by andersonbarutti
Get week number
获取周数
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());
回答by Nico M
I know this is an old question, but maybe it helps:
我知道这是一个老问题，但也许它有帮助：
回答by S.Younes
/*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)));