I'm using this javascript to check if the age entered is older than 18.
function calculateDiffYear(date, month, year)
{
var cur = new Date();
var diff = Math.floor((cur.getTime() - new Date(year, month, date)) / (60 * 60 * 24 * 1000));
diff -= Math.floor((cur.getFullYear() - year) / 4);
return diff / 365;
}
function checkBorn(sender)
{
var root = sender.form;
var date = root.elements['date'].value;
var month = root.elements['month'].value - 1;
var year = root.elements['year'].value;
if (!isValidDate(date, month, year) || calculateDiffYear(date, month, year) < 18) return false;
return true;
}
If works almost right, except for, if we are in a leap year, it gives older than 18 to a person who becomes 18 tomorrow, at least in the tests I'm doing with today date and changing to las year. I tryed adding this but no luck:
if ($('#leap').val()) divider = 366;
else divider = 365;
return diff / divider;
Do you know how can I solve it?
Thank you
If I wanted to test if a particular date was more than 18 years ago I'd do something like this:
function meetsMinimumAge(birthDate, minAge) {
var tempDate = new Date(birthDate.getFullYear() + minAge, birthDate.getMonth(), birthDate.getDate());
return (tempDate <= new Date());
}
if (meetsMinimumAge(new Date(year, month, date), 18)) {
// is OK, do something
} else {
// too young - error
}
Essentially this takes the supplied birthday, adds 18 to it, and checks if that is still on or before today's date.
My age-checking code goes something like this:
function checkAge(dateofbirth) {
var yd, md, dd, now = new Date();
yd = now.getUTCFullYear()-dateofbirth.getUTCFullYear();
md = now.getUTCMonth()-dateofbirth.getUTCMonth();
dd = now.getUTCDate()-dateofbirth.getUTCDate();
if( yd > 18) return true;
if( md > 0) return true;
return dd >= 0;
}
Basically, if the year difference is 19 or more, then they must be over 18.
Otherwise, if the current month is past the month of birth, they are 18 and a few months old.
Otherwise, if the current day is greater than or equal to the day of birth, they are 18 and a few days old (or it is their 18th birthday).
This works regardless of leap years and is much more efficient than your current code.
You can use moment.js to validate it:
var yourDate = year.toString() + "/" + month.toString() + "/" day.toString();
var date = moment(yourDate, "YYYY/MM/DD"); // There are other formats!
var years = moment().diff(date, 'years', false);
if(years >= 18){
return true;
}
return false;
Related
This question already has answers here:
How to calculate number of days between two dates?
(42 answers)
Closed last month.
I am new to javascript.
I have specific month columns (9/30/2022, 10/31/2022,11/30/2022). I have a contract with a start date and end date (spanning multiple months).
I need to determine the number of days the contract was active for a specific month column.
Example:
Contact Start Date: 09/15/2022
Contract End Date: 10/24/2022
Number of days the contract was active in Sept 2022 is 16.
I found the code below that gives me the contract period broken down for each month (i.e.) **9 - 17; 10 - 23; **
Thank you in advance for any assistance.
I found this code
function getDays() {
var dropdt = new Date(document.getElementById("arr").value);
var pickdt = new Date(document.getElementById("dep").value);
var result = "";
for (var year = dropdt.getFullYear(); year <= pickdt.getFullYear(); year++) {
var firstMonth = (year == dropdt.getFullYear()) ? dropdt.getMonth() : 0;
var lastMonth = (year == pickdt.getFullYear()) ? pickdt.getMonth() : 11;
for (var month = firstMonth; month <= lastMonth; month++) {
var firstDay = (year === dropdt.getFullYear() && month === firstMonth) ? dropdt.getDate() : 1;
var lastDay = (year === pickdt.getFullYear() && month === lastMonth) ? pickdt.getDate() : 0;
var lastDateMonth = (lastDay === 0) ? (month + 1) : month
var firstDate = new Date(year, month, firstDay);
var lastDate = new Date(year, lastDateMonth, lastDay);
result += (month + 1) + " - " + parseInt((lastDate - firstDate) / (24 * 3600 * 1000) + 1) + "; ";
}
}
return result;
}
function cal() {
if (document.getElementById("dep")) {
document.getElementById("number-of-dates").value = getDays();
}
Calculate
`
The following snippet will generate an object res with the keys being zero-based month-indexes ("8" is for September, "9" for October, etc.) and the values are the number of days for each of these months:
const start=new Date("2022-09-15");
const end=new Date("2022-11-24");
let nextFirst=new Date(start.getTime()), month, days={};
do {
month=nextFirst.getMonth();
nextFirst.setMonth(month+1);nextFirst.setDate(1);
days[month]=Math.round(((nextFirst<end?nextFirst:end)-start)/86400000);
start.setTime(nextFirst.getTime());
} while(nextFirst<end)
console.log(days);
This can be extended into a more reliable function returning a year-month combination:
function daysPerMonth(start,end){
let nextFirst=new Date(start.getTime()), year, month, days={};
do {
year=nextFirst.getFullYear();
month=nextFirst.getMonth();
nextFirst.setMonth(month+1);nextFirst.setDate(1);
days[`${year}-${String(1+month).padStart(2,"0")}`]=Math.round(((nextFirst<end?nextFirst:end)-start)/86400000);
start.setTime(nextFirst.getTime());
} while(nextFirst<end)
return days;
}
[["2022-09-15","2022-11-24"],["2022-11-29","2023-02-04"]].forEach(([a,b])=>
console.log(daysPerMonth(new Date(a),new Date(b))));
Managing and calculating dates, times, and date-times are notoriously finicky in javascript and across browsers. Rather than trying to define your own logic for this use the famous Moment.js library.
In particular to calculate the length between two dates you can utilize the diff function between two moments
var a = moment([2007, 0, 29]);
var b = moment([2007, 0, 25]);
var diff = a.diff(b, 'days') // 1
console.log(diff);
<script src="https://momentjs.com/downloads/moment.js"></script>
The supported measurements are years, months, weeks, days, hours, minutes, and seconds.
This question already has an answer here:
How to take the differences between two dates, depending on the date of birth [duplicate]
(1 answer)
Closed 2 years ago.
function test(val) {
year = parseInt(val.slice(0,2)); // get year
month = parseInt(val.slice(2,4)); // get month
date = val.slice(4,6); // get date
if (month > 40) { // For people born after 2000, 40 is added to the month.
year += 2000;
month -= 40;
} else {
year += 1900;
}
date = new Date(year, month-1, date, 0, 0);
date_now = new Date();
var diff =(date_now.getTime() - date.getTime()) / 1000;
diff /= (60 * 60 * 24);
diff = Math.abs(Math.floor(diff/365.25));
console.log(diff);
}
test("940911") // Should return 25
test("940910") // Should return 26
test("940909") // Should return 26
So If I'm born 1994.09.11 function should returns 25, Because 09.11 is tommorrow, but if I'm born at 1994.09.09, It should returns 26, because 09.09 was yesterday. I do not want to use libreries like moment.js etc.
Here the solution, anyway I suggest you to use moment.
So what about this solution?
Anyway the year? If 01 is 1901 or 2001 ?
function test(d){
let year = 1900+parseInt(d.slice(0,2));
let month = parseInt(d.slice(2,4));
let day = d.slice(4,6);
let yearNow = new Date().getFullYear();
let monthNow = new Date().getMonth() + 1;
let dayNow = new Date().getDate();
if (monthNow === month && dayNow < day || monthNow < month) {
return yearNow - year - 1;
} else {
return yearNow - year;
}
}
console.log(test("940911"))
console.log(test("940910"))
console.log(test("940909"))
I have two sets of codes that work. Needed help combining them into one.
This code gets me the difference between two dates. works perfectly:
function test(){
var date1 = new Date(txtbox_1.value);
var date2 = new Date(txtbox_2.value);
var diff = (date2 - date1)/1000;
var diff = Math.abs(Math.floor(diff));
var days = Math.floor(diff/(24*60*60));
var leftSec = diff - days * 24*60*60;
var hrs = Math.floor(leftSec/(60*60));
var leftSec = leftSec - hrs * 60*60;
var min = Math.floor(leftSec/(60));
var leftSec = leftSec - min * 60;
txtbox_3.value = days + "." + hrs; }
source for the above code
The code below by #cyberfly appears to have the answer of excluding sat and sun which is what i needed. source. However, its in jquery and the above code is in JS. Therefore, needed help combining as i lacked that knowledge :(
<script type="text/javascript">
$("#startdate, #enddate").change(function() {
var d1 = $("#startdate").val();
var d2 = $("#enddate").val();
var minutes = 1000*60;
var hours = minutes*60;
var day = hours*24;
var startdate1 = getDateFromFormat(d1, "d-m-y");
var enddate1 = getDateFromFormat(d2, "d-m-y");
var days = calcBusinessDays(new Date(startdate1),new Date(enddate1));
if(days>0)
{ $("#noofdays").val(days);}
else
{ $("#noofdays").val(0);}
});
</script>
EDIT
Made an attempt at combining the codes. here is my sample. getting object expected error.
function test(){
var date1 = new Date(startdate.value);
var date2 = new Date(enddate.value);
var diff = (date2 - date1)/1000;
var diff = Math.abs(Math.floor(diff));
var days = Math.floor(diff/(24*60*60));
var leftSec = diff - days * 24*60*60;
var hrs = Math.floor(leftSec/(60*60));
var leftSec = leftSec - hrs * 60*60;
var min = Math.floor(leftSec/(60));
var leftSec = leftSec - min * 60;
var startdate1 = getDateFromFormat(startdate, "dd/mm/yyyy hh:mm");
var enddate1 = getDateFromFormat(enddate, "dd/mm/yyyy hh:mm");
days = calcBusinessDays(new Date(startdate1),new Date(enddate1));
noofdays.value = days + "." + hrs; }
start: <input type="text" id="startdate" name="startdate" value="02/03/2015 00:00">
end: <input type="text" id="enddate" name="enddate" value="02/03/2015 00:01">
<input type="text" id="noofdays" name="noofdays" value="">
When determining the number of days between two dates, there are lots of decisions to be made about what is a day. For example, the period 1 Feb to 2 Feb is generally one day, so 1 Feb to 1 Feb is zero days.
When adding the complexity of counting only business days, things get a lot tougher. E.g. Monday 2 Feb 2015 to Friday 6 February is 4 elapsed days (Monday to Tuesday is 1, Monday to Wednesday is 2, etc.), however the expression "Monday to Friday" is generally viewed as 5 business days and the duration Mon 2 Feb to Sat 7 Feb should also be 4 business days, but Sunday to Saturday should be 5.
So here's my algorithm:
Get the total number of whole days between the two dates
Divide by 7 to get the number of whole weeks
Multiply the number of weeks by two to get the number of weekend days
Subtract the number of weekend days from the whole to get business days
If the number of total days is not an even number of weeks, add the numbe of weeks * 7 to the start date to get a temp date
While the temp date is less than the end date:
if the temp date is not a Saturday or Sunday, add one the business days
add one to the temp date
That's it.
The stepping part at the end can probably be replaced by some other algorithm, but it will never loop for more than 6 days so it's a simple and reasonably efficient solution to the issue of uneven weeks.
Some consequences of the above:
Monday to Friday is 4 business days
Any day to the same day in a different week is an even number of weeks and therefore an even mutiple of 5, e.g. Monday 2 Feb to Monday 9 Feb and Sunday 1 Feb to Sunday 8 Feb are 5 business days
Friday 6 Feb to Sunday 7 Feb is zero business days
Friday 6 Feb to Monday 9 Feb is one business day
Sunday 8 Feb to: Sunday 15 Feb, Sat 14 Feb and Fri 13 Feb are all 5 business days
Here's the code:
// Expects start date to be before end date
// start and end are Date objects
function dateDifference(start, end) {
// Copy date objects so don't modify originals
var s = new Date(+start);
var e = new Date(+end);
// Set time to midday to avoid dalight saving and browser quirks
s.setHours(12,0,0,0);
e.setHours(12,0,0,0);
// Get the difference in whole days
var totalDays = Math.round((e - s) / 8.64e7);
// Get the difference in whole weeks
var wholeWeeks = totalDays / 7 | 0;
// Estimate business days as number of whole weeks * 5
var days = wholeWeeks * 5;
// If not even number of weeks, calc remaining weekend days
if (totalDays % 7) {
s.setDate(s.getDate() + wholeWeeks * 7);
while (s < e) {
s.setDate(s.getDate() + 1);
// If day isn't a Sunday or Saturday, add to business days
if (s.getDay() != 0 && s.getDay() != 6) {
++days;
}
}
}
return days;
}
I don't know how it compares to jfriend00's answer or the code you referenced, if you want the period to be inclusive, just add one if the start or end date are a business day.
Here's a simple function to calculate the number of business days between two date objects. As designed, it does not count the start day, but does count the end day so if you give it a date on a Tuesday of one week and a Tuesday of the next week, it will return 5 business days. This does not account for holidays, but does work properly across daylight savings changes.
function calcBusinessDays(start, end) {
// This makes no effort to account for holidays
// Counts end day, does not count start day
// make copies we can normalize without changing passed in objects
var start = new Date(start);
var end = new Date(end);
// initial total
var totalBusinessDays = 0;
// normalize both start and end to beginning of the day
start.setHours(0,0,0,0);
end.setHours(0,0,0,0);
var current = new Date(start);
current.setDate(current.getDate() + 1);
var day;
// loop through each day, checking
while (current <= end) {
day = current.getDay();
if (day >= 1 && day <= 5) {
++totalBusinessDays;
}
current.setDate(current.getDate() + 1);
}
return totalBusinessDays;
}
And, the jQuery + jQueryUI code for a demo:
// make both input fields into date pickers
$("#startDate, #endDate").datepicker();
// process click to calculate the difference between the two days
$("#calc").click(function(e) {
var diff = calcBusinessDays(
$("#startDate").datepicker("getDate"),
$("#endDate").datepicker("getDate")
);
$("#diff").html(diff);
});
And, here's a simple demo built with the date picker in jQueryUI: http://jsfiddle.net/jfriend00/z1txs10d/
const firstDate = new Date("December 30, 2020");
const secondDate = new Date("January 4, 2021");
const daysWithOutWeekEnd = [];
for (var currentDate = new Date(firstDate); currentDate <= secondDate; currentDate.setDate(currentDate.getDate() + 1)) {
// console.log(currentDate);
if (currentDate.getDay() != 0 && currentDate.getDay() != 6) {
daysWithOutWeekEnd.push(new Date(currentDate));
}
}
console.log(daysWithOutWeekEnd, daysWithOutWeekEnd.length);
#RobG has given an excellent algorithm to separate business days from weekends.
I think the only problem is if the starting days is a weekend, Saturday or Sunday, then the no of working days/weekends will one less.
Corrected code is below.
function dateDifference(start, end) {
// Copy date objects so don't modify originals
var s = new Date(start);
var e = new Date(end);
var addOneMoreDay = 0;
if( s.getDay() == 0 || s.getDay() == 6 ) {
addOneMoreDay = 1;
}
// Set time to midday to avoid dalight saving and browser quirks
s.setHours(12,0,0,0);
e.setHours(12,0,0,0);
// Get the difference in whole days
var totalDays = Math.round((e - s) / 8.64e7);
// Get the difference in whole weeks
var wholeWeeks = totalDays / 7 | 0;
// Estimate business days as number of whole weeks * 5
var days = wholeWeeks * 5;
// If not even number of weeks, calc remaining weekend days
if (totalDays % 7) {
s.setDate(s.getDate() + wholeWeeks * 7);
while (s < e) {
s.setDate(s.getDate() + 1);
// If day isn't a Sunday or Saturday, add to business days
if (s.getDay() != 0 && s.getDay() != 6) {
++days;
}
//s.setDate(s.getDate() + 1);
}
}
var weekEndDays = totalDays - days + addOneMoreDay;
return weekEndDays;
}
JSFiddle link is https://jsfiddle.net/ykxj4k09/2/
First Get the Number of Days in a month
totalDays(month, year) {
return new Date(year, month, 0).getDate();
}
Then Get No Of Working Days In A Month By removing Saturday and Sunday
totalWorkdays() {
var d = new Date(); // to know present date
var m = d.getMonth() + 1; // to know present month
var y = d.getFullYear(); // to knoow present year
var td = this.totalDays(m, y);// to get no of days in a month
for (var i = 1; i <= td; i++) {
var s = new Date(y, m - 1, i);
if (s.getDay() != 0 && s.getDay() != 6) {
this.workDays.push(s.getDate());// working days
}else {
this.totalWeekDays.push(s.getDate());//week days
}
}
this.totalWorkingDays = this.workDays.length;
}
I thought the above code snippets others shared are lengthy.
I am sharing a concise snippet that gives date after considering the total number of days specified. we can also customize dates other than Saturdays and Sundays.
function getBusinessDays(dateObj, days) {
for (var i = 0; i < days; i++) {
if (days > 0) {
switch (dateObj.getDay()) {
// 6 being Saturday and 0 being Sunday.
case 6, 0:
dateObj.setDate(dateObj.getDate() + 2)
break;
//5 = Friday.
case 5:
dateObj.setDate(dateObj.getDate() + 3)
break;
//handle Monday, Tuesday, Wednesday and Thursday!
default:
dateObj.setDate(dateObj.getDate() + 1)
//console.log(dateObj)
break;
}
}
}
return dateObj;
}
console.log(getBusinessDays(new Date(), 11))
//Mon Dec 20 2021 18:56:01 GMT+0530 (India Standard Time)
Hey javascript masters,
Attempting to create an age verification page to a client's site. Code below is not functioning as it doesn't matter what year you select, it will still allow you to enter the site. Not sure what I should be looking at to correct.
Any help is appreciated.
<script type="text/javascript"><!--
function checkAge(f){
var dob=new Date();
var date=dob.getDate();
var month=dob.getMonth() + 1;
var year=dob.getFullYear();
var cmbmonth=parseInt(document.getElementById("cmbmonth").options[document.getElementById("cmbmonth").selectedIndex].value);
var cmbday=parseInt(document.getElementById("cmbday").options[document.getElementById("cmbday").selectedIndex].value);
var cmbyear=parseInt(document.getElementById("cmbyear").options[document.getElementById("cmbyear").selectedIndex].value);
age=year-cmbyear;
if(cmbmonth>month){age--;}
else{if(cmbmonth==month && cmbday>=date){age--;}}
if(cmbmonth==0){alert("You must enter the month you were born in.");return false;}
else if(cmbday==0){alert("You must enter the day you were born on.");return false;}
else if(cmbyear==2005){alert("You must enter the year you were born in.");return false;}
else if(age<13){alert("You are unable to view this site!");location.replace("http://www.dharmatalks.org");return false;}
else{return true;}
}
// --></script>
Calculating age in years, months and days is a bit trickier than it should be due to the differences in month and year lengths. Here's a function that will return the difference between two dates in years, months, days, hours, minutes and seconds.
function dateDifference(start, end) {
// Copy date objects so don't modify originals
var s = new Date(+start);
var e = new Date(+end);
var timeDiff, years, months, days, hours, minutes, seconds;
// Get estimate of year difference
years = e.getFullYear() - s.getFullYear();
// Add difference to start, if greater than end, remove one year
// Note start from restored start date as adding and subtracting years
// may not be symetric
s.setFullYear(s.getFullYear() + years);
if (s > e) {
--years;
s = new Date(+start);
s.setFullYear(s.getFullYear() + years);
}
// Get estimate of months
months = e.getMonth() - s.getMonth();
months += months < 0? 12 : 0;
// Add difference to start, adjust if greater
s.setMonth(s.getMonth() + months);
if (s > e) {
--months;
s = new Date(+start);
s.setFullYear(s.getFullYear() + years);
s.setMonth(s.getMonth() + months);
}
// Get remaining time difference, round to next full second
timeDiff = (e - s + 999) / 1e3 | 0;
days = timeDiff / 8.64e4 | 0;
hours = (timeDiff % 8.64e4) / 3.6e3 | 0;
minutes = (timeDiff % 3.6e3) / 6e1 | 0;
seconds = timeDiff % 6e1;
return [years, months, days, hours, minutes, seconds];
}
You can abbreviate the above just after the year part and return just that if you want.
Note that in your code:
var cmbmonth=parseInt(document.getElementById("cmbmonth").options[document.getElementById("cmbmonth").selectedIndex].value);
can be:
var cmbmonth = document.getElementById("cmbmonth").value;
There is no need for parseInt, the Date constructor will happily work with string values. If you have used calendar month numbers for the values (i.e. Jan = 1) then subtract 1 before giving it to the Date constructor, but simpler to use javascript month indexes for the values (i.e. Jan = 0).
You can then do:
var diff = dateDifference(new Date(cmbyear, cmbmonth, cmbdate), new Date());
if (diff[0] < 18) {
// sorry, under 18
}
This question already has answers here:
Get week of year in JavaScript like in PHP
(23 answers)
Closed 5 years ago.
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;
}
/**
* 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) == 'number' ? 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
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);
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)
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());
I know this is an old question, but maybe it helps:
http://weeknumber.net/how-to/javascript
// This script is released to the public domain and may be used, modified and
// distributed without restrictions. Attribution not necessary but appreciated.
// Source: https://weeknumber.net/how-to/javascript
// Returns the ISO week of the date.
Date.prototype.getWeek = function() {
var date = new Date(this.getTime());
date.setHours(0, 0, 0, 0);
// Thursday in current week decides the year.
date.setDate(date.getDate() + 3 - (date.getDay() + 6) % 7);
// January 4 is always in week 1.
var week1 = new Date(date.getFullYear(), 0, 4);
// Adjust to Thursday in week 1 and count number of weeks from date to week1.
return 1 + Math.round(((date.getTime() - week1.getTime()) / 86400000
- 3 + (week1.getDay() + 6) % 7) / 7);
}
// Returns the four-digit year corresponding to the ISO week of the date.
Date.prototype.getWeekYear = function() {
var date = new Date(this.getTime());
date.setDate(date.getDate() + 3 - (date.getDay() + 6) % 7);
return date.getFullYear();
}
/*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)));