JavaScript calculate the day of the year 1 - 366

Question

How do I use JavaScript to calculate the day of the year  from 1 - 366  For example   January 3 should be 3  February 1 should be 32

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This works across Daylight Savings Time changes in all countries  the  noon  one above doesn t work in Australia    Date prototype isLeapYear   function         var year   this getFullYear        if  year  amp  3     0  return false      return   year   100     0     year   400     0          Get Day of Year Date prototype getDOY   function         var dayCount    0  31  59  90  120  151  181  212  243  273  304  334       var mn   this getMonth        var dn   this getDate        var dayOfYear   dayCount mn    dn      if mn  gt  1  amp  amp  this isLeapYear    dayOfYear        return dayOfYear

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Math round  new Date   setHours 23  - new Date new Date   getFullYear    0  1  0  0  0   1000 86400    further optimizes the answer   Moreover  by changing setHours 23  or the last-but-two zero later on to another value may provide day-of-year related to another timezone  For example  to retrieve from Europe a resource located in America

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I think this is more straightforward    var date365   0   var currentDate   new Date    var currentYear   currentDate getFullYear    var currentMonth   currentDate getMonth     var currentDay   currentDate getDate      var monthLength    31 28 31 30 31 30 31 31 30 31 30 31    var leapYear   new Date currentYear  1  29    if  leapYear getDate      29       If it s a leap year  changes 28 to 29     monthLength 1    29     for   i 0  i  lt  currentMonth  i            date365   date365   monthLength i     date365   date365   currentDay     Done

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This method takes into account timezone issue and daylight saving time  function dayofyear d         d is a Date object     var yn   d getFullYear        var mn   d getMonth        var dn   d getDate        var d1   new Date yn 0 1 12 0 0      noon on Jan  1     var d2   new Date yn mn dn 12 0 0      noon on input date     var ddiff   Math round  d2-d1  864e5       return ddiff 1        took from here   See also this fiddle

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I would like to provide a solution that does calculations adding the days for each previous month    x000D   x000D  function getDayOfYear date    x000D      var month   date getMonth    x000D      var year   date getFullYear    x000D      var days   date getDate    x000D      for  var i   0  i  lt  month  i      x000D          days    new Date year  i 1  0  getDate    x000D        x000D      return days  x000D    x000D  var input   new Date 2017  7  5   x000D  console log input   x000D  console log getDayOfYear input    x000D   x000D   x000D    This way you don t have to manage the details of leap years and daylight saving

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Luckily this question doesn t specify if the number of the current day is required  leaving room for this answer  Also some answers  also on other questions  had leap-year problems or used the Date-object  Although javascript s Date object covers approximately 285616 years  100 000 000 days  on either side of January 1 1970  I was fed up with all kinds of unexpected date inconsistencies across different browsers  most notably year 0 to 99   I was also curious how to calculate it   So I wrote a simple and above all  small algorithm to calculate the correct  Proleptic Gregorian   Astronomical   ISO 8601 2004  clause 4 3 2 1   so year 0 exists and is a leap year and negative years are supported  day of the year based on year  month and day  Note that in AD BC notation  year 0 AD BC does not exist  instead year 1 BC is the leap-year  IF you need to account for BC notation then simply subtract one year of the  otherwise positive  year-value first    I modified  for javascript  the short-circuit bitmask-modulo leapYear algorithm and came up with a magic number to do a bit-wise lookup of offsets  that excludes jan and feb  thus needing 10   3 bits  30 bits is less than 31 bits  so we can safely save another character on the bitshift instead of  gt  gt  gt      Note that neither month or day may be 0  That means that if you need this equation just for the current day  feeding it using  getMonth    you just need to remove the -- from --m   Note this assumes a valid date  although error-checking is just some characters more     x000D   x000D  function dayNo y m d   x000D    return --m 31- m gt 1  1054267675 gt  gt m 3-6 amp 7 - y amp 3    y 25  amp  amp y amp 15 0 1  0  d  x000D    x000D   lt  -- some examples for the snippet -- gt  x000D   lt input type text value   - Y-M-D  onblur   x000D    var d this value match   -  d     d    d d     d    d d          x000D    this nextSibling innerHTML   Day      dayNo  d 1    d 2    d 3    x000D      gt  lt span gt  lt  span gt  x000D   x000D   lt br gt  lt hr gt  lt br gt  x000D   x000D   lt button onclick   x000D    var d new Date    x000D    this nextSibling innerHTML dayNo d getFullYear    d getMonth   1  d getDate        Day s    x000D    gt get current dayno  lt  button gt  lt span gt  lt  span gt  x000D   x000D   x000D      Here is the version with correct range-validation         x000D   x000D  function dayNo y m d   x000D    return --m gt  0  amp  amp  m lt 12  amp  amp  d gt 0  amp  amp  d lt 29     x000D             4  y y amp 3    y 25  amp  amp y amp 15 0 1  15662003 gt  gt m 2 amp 3   x000D              amp  amp  m 31- m gt 1  1054267675 gt  gt m 3-6 amp 7 -y 0  d  x000D    x000D   lt  -- some examples for the snippet -- gt  x000D   lt input type text value   - Y-M-D  onblur   x000D    var d this value match   -  d     d    d d     d    d d          x000D    this nextSibling innerHTML   Day      dayNo  d 1    d 2    d 3    x000D      gt  lt span gt  lt  span gt  x000D   x000D   x000D    Again  one line  but I split it into 3 lines for readability  and following explanation         The last line is identical to the function above  however the  identical  leapYear algorithm is moved to a previous short-circuit section  before the day-number calculation   because it is also needed to know how much days a month has in a given  leap  year        The middle line calculates the correct offset number  for max number of days  for a given month in a given  leap year using another magic number  since 31-28 3 and 3 is just 2 bits  then 12 2 24 bits  we can store all 12 months  Since addition can be faster then subtraction  we add the offset  instead of subtract it from 31   To avoid a leap-year decision-branch for February  we modify that magic lookup-number on the fly   That leaves us with the  pretty obvious  first line  it checks that month and date are within valid bounds and ensures us with a false return value on range error  note that this function also should not be able to return 0  because 1 jan 0000 is still day 1    providing easy error-checking  if r dayNo   y  m  d        If used this way  where month and day may not be 0   then one can change --m gt  0  amp  amp  m lt 12 to m gt 0  amp  amp  --m lt 12  saving another char   The reason I typed the snippet in it s current form is that for 0-based month values  one just needs to remove the -- from --m   Extra  Note  don t use this day s per month algorithm if you need just max day s per month  In that case there is a more efficient algorithm  because we only need leepYear when the month is February  I posted as answer this question  What is the best way to determine the number of days in a month with javascript

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A straightforward solution with complete explanation   x000D   x000D  var dayOfYear   function date      const daysInMonth    31  28  31  30  31  30  31  31  30  31  30  31     const  yyyy  mm  dd    date split  -   map Number         Checks if February has 29 days   const isLeap    year    gt  new Date year  1  29  getDate       29        If it s a leap year  changes 28 to 29   if  isLeap yyyy   daysInMonth 1    29     let daysBeforeMonth   0       Slice the array and exclude the current Month   for  const i of daysInMonth slice 0  mm - 1         daysBeforeMonth    i         return daysBeforeMonth   dd      console log dayOfYear  2020-1-3     console log dayOfYear  2020-2-1     x000D   x000D   x000D

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Date prototype dayOfYear  function        var j1  new Date this       j1 setMonth 0  0       return Math round  this-j1  8 64e7      alert new Date   dayOfYear

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Math floor  Date now   - Date parse new Date   getFullYear    0  0     86400000    this is my solution

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It always get s me worried when mixing maths with date functions  it s so easy to miss some leap year other detail   Say you have   var d   new Date      I would suggest using the following  days will be saved in day   for var day   d getDate    d getMonth    day    d getDate        d setDate 0     Can t see any reason why this wouldn t work just fine  and I wouldn t be so worried about the few iterations since this will not be used so intensively

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Well  if I understand you correctly  you want 366 on a leap year  365 otherwise  right  A year is a leap year if it s evenly divisible by 4 but not by 100 unless it s also divisible by 400  function daysInYear year        if year   4     0  amp  amp   year   100     0    year   400     0                Leap year         return 366        else              Not a leap year         return 365            Edit after update  In that case  I don t think there s a built-in method  you ll need to do this  function daysInFebruary year        if year   4     0  amp  amp   year   100     0    year   400     0                Leap year         return 29        else              Not a leap year         return 28           function dateToDay date        var feb   daysInFebruary date getFullYear         var aggregateMonths    0     January                            31     February                            31   feb     March                            31   feb   31     April                            31   feb   31   30     May                            31   feb   31   30   31     June                            31   feb   31   30   31   30     July                            31   feb   31   30   31   30   31     August                            31   feb   31   30   31   30   31   31     September                            31   feb   31   30   31   30   31   31   30     October                            31   feb   31   30   31   30   31   31   30   31     November                            31   feb   31   30   31   30   31   31   30   31   30     December                                 return aggregateMonths date getMonth      date getDate        Yes  I actually did that without copying or pasting  If there s an easy way I ll be mad

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This is a solution that avoids the troublesome Date object and timezone issues  it requires that your input date be in the format  yyyy-dd-mm   If you want to change the format  then modify date str to parts function       function get day of year str date       var date parts   date str to parts str date       var is leap    date parts year 4   0      var acct for leap    is leap  amp  amp  date parts month gt 2       var day of year   0       var ary months             0          31    jan         28    feb non leap          31    march         30    april         31    may         30    june         31    july         31    aug         30    sep         31    oct         30    nov            31    dec                  for var i 1  i  lt  date parts month  i             day of year    ary months i              day of year    date parts date       if  acct for leap   day of year  1       return day of year      function date str to parts str date       return            year  parseInt str date substr 0 4  10            month  parseInt str date substr 5 2  10            date  parseInt str date substr 8 2  10

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For those among us who want a fast alternative solution    x000D   x000D   function    use strict   x000D  function daysIntoTheYear dateInput   x000D      var fullYear   dateInput getFullYear   0  x000D       Leap Years are any year that can be exactly divided by 4  2012  2016  etc  x000D       except if it can be exactly divided by 100  then it isn t  2100  2200  etc  x000D        except if it can be exactly divided by 400  then it is  2000  2400   x000D       https   www mathsisfun com leap-years html   x000D      var isLeapYear     fullYear  amp  3     fullYear 100  amp  3       0   1   0  x000D       fullYear  amp  3     fullYear   4   but faster x000D        Alternative var isLeapYear  new Date currentYear 1 29 12   getDate     29 1 0 x000D      var fullMonth   dateInput getMonth   0  x000D      return    x000D             Calculate the day of the year in the Gregorian calendar x000D             The code below works based upon the facts of signed right shifts x000D                     x   gt  gt  n  shifts n and fills in the n highest bits with 0s  x000D                     -x   gt  gt  n  shifts n and fills in the n highest bits with 1s x000D              This assumes that x is a positive integer  x000D           31  amp    -fullMonth   gt  gt  4        January     -11  gt  gt 4   -1 x000D            28   isLeapYear   amp    1-fullMonth   gt  gt  4        February x000D           31  amp    2-fullMonth   gt  gt  4        March x000D           30  amp    3-fullMonth   gt  gt  4        April x000D           31  amp    4-fullMonth   gt  gt  4        May x000D           30  amp    5-fullMonth   gt  gt  4        June x000D           31  amp    6-fullMonth   gt  gt  4        July x000D           31  amp    7-fullMonth   gt  gt  4        August x000D           30  amp    8-fullMonth   gt  gt  4        September x000D           31  amp    9-fullMonth   gt  gt  4        October x000D           30  amp    10-fullMonth   gt  gt  4        November x000D             There are no months past December  the year rolls into the next  x000D             Thus  fullMonth is 0-based  so it will never be 12 in Javascript x000D   x000D           dateInput getDate   0     get day of the month x000D   x000D        amp 0xffff   x000D    x000D     Demonstration  x000D  var date   new Date 2100  0  1  x000D  for  var i 0  i lt 12  i i 1 0  date setMonth date getMonth   1 0   x000D      console log date getMonth      tday   daysIntoTheYear date    t  date   x000D  date   new Date 1900  0  1   x000D  for  var i 0  i lt 12  i i 1 0  date setMonth date getMonth   1 0   x000D      console log date getMonth      tday   daysIntoTheYear date    t  date   x000D   x000D     Performance Benchmark  x000D  console time  Speed of processing 65536 dates    x000D  for  var i 0 month date getMonth   0  i lt 65536  i i 1 0  x000D      date setMonth month month 1  daysIntoTheYear date  0  0   x000D  console timeEnd  Speed of processing 65536 dates    x000D        x000D   x000D   x000D    The size of the months of the year and the way that Leap Years work fits perfectly into keeping our time on track with the sun  Heck  it works so perfectly that all we ever do is just adjust mere seconds here and there  Our current system of leap years has been in effect since February 24th  1582  and will likely stay in effect for the foreseeable future   DST  however  is very subject to change  It may be that 20 years from now  some country may offset time by a whole day or some other extreme for DST  A whole DST day will almost certainly never happen  but DST is still nevertheless very up-in-the-air and indecisive  Thus  the above solution is future proof in addition to being very very fast   The above code snippet runs very fast  My computer can process 65536 dates in  52ms on Chrome

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I find it very interesting that no one considered using UTC since it is not subject to DST  Therefore  I propose the following   function daysIntoYear date       return  Date UTC date getFullYear    date getMonth    date getDate    - Date UTC date getFullYear    0  0     24   60   60   1000      You can test it with the following    new Date 2016 0 1   new Date 2016 1 1   new Date 2016 2 1   new Date 2016 5 1   new Date 2016 11 31        forEach d   gt           console log    d toLocaleDateString    is   daysIntoYear d   days into the year       Which outputs for the leap year 2016  verified using http   www epochconverter com days 2016    1 1 2016 is 1 days into the year 2 1 2016 is 32 days into the year 3 1 2016 is 61 days into the year 6 1 2016 is 153 days into the year 12 31 2016 is 366 days into the year

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I ve made one that s readable and will do the trick very quickly  as well as handle JS Date objects with disparate time zones   I ve included quite a few test cases for time zones  DST  leap seconds and Leap years   P S  ECMA-262 ignores leap seconds  unlike UTC  If you were to convert this to a language that uses real UTC  you could just add 1 to oneDay    x000D   x000D     returns 1 - 366 x000D  findDayOfYear   function  date    x000D    var oneDay   1000   60   60   24     A day in milliseconds x000D    var og                               Saving original data x000D      ts  date getTime    x000D      dom  date getDate                  We don t need to save hours minutes because DST is never at 12am  x000D      month  date getMonth   x000D      x000D    date setDate 1                       Sets Date of the Month to the 1st  x000D    date setMonth 0                      Months are zero based in JS s Date object x000D    var start ts   date getTime          New Year s Midnight JS Timestamp x000D    var diff   og ts - start ts  x000D   x000D    date setDate og dom                  Revert back to original date object x000D    date setMonth og month               This method does preserve timezone x000D    return Math round diff   oneDay    1     Deals with DST globally  Ceil fails in Australia  Floor Fails in US  x000D    x000D   x000D     Tests x000D  var pre start dst   new Date 2016  2  12   x000D  var on start dst   new Date 2016  2  13   x000D  var post start dst   new Date 2016  2  14   x000D   x000D  var pre end dst date   new Date 2016  10  5   x000D  var on end dst date   new Date 2016  10  6   x000D  var post end dst date   new Date 2016  10  7   x000D   x000D  var pre leap second   new Date 2015  5  29   x000D  var on leap second   new Date 2015  5  30   x000D  var post leap second   new Date 2015  6  1   x000D   x000D     2012 was a leap year with a leap second in june 30th x000D  var leap second december31 premidnight   new Date 2012  11  31  23  59  59  999   x000D   x000D  var january1   new Date 2016  0  1   x000D  var january31   new Date 2016  0  31   x000D   x000D  var december31   new Date 2015  11  31   x000D  var leap december31   new Date 2016  11  31   x000D   x000D  alert     x000D        nPre Start DST      findDayOfYear pre start dst          72  x000D        nOn Start DST      findDayOfYear on start dst          73  x000D        nPost Start DST      findDayOfYear post start dst          74  x000D         x000D        nPre Leap Second      findDayOfYear pre leap second          180  x000D        nOn Leap Second      findDayOfYear on leap second          181  x000D        nPost Leap Second      findDayOfYear post leap second          182  x000D         x000D        nPre End DST      findDayOfYear pre end dst date          310  x000D        nOn End DST      findDayOfYear on end dst date          311  x000D        nPost End DST      findDayOfYear post end dst date          312  x000D         x000D        nJanuary 1st      findDayOfYear january1          1  x000D        nJanuary 31st      findDayOfYear january31          31  x000D        nNormal December 31st      findDayOfYear december31          365  x000D        nLeap December 31st      findDayOfYear leap december31          366  x000D        nLast Second of Double Leap      findDayOfYear leap second december31 premidnight          366  x000D     x000D   x000D   x000D

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A alternative using UTC timestamps  Also as others noted the day indicating 1st a month is 1 rather than 0  The month starts at 0 however    var now   Date now    var year    new Date   getUTCFullYear    var year start   Date UTC year  0  1   var day length in ms   1000 60 60 24  var day number   Math floor  now - year start  day length in ms  console log  Day of year     day number

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I wrote these two javascript functions which return the day of the year  Jan 1   1   Both of them account for leap years  function dayOfTheYear        for today var M  31 28 31 30 31 30 31 31 30 31 30 31   var x new Date    var m x getMonth    var y x getFullYear    if  y   400    0     y   4    0  amp  amp  y   100    0      M 1    var Y 0  for  var i 0 i lt m   i   Y  M i    return Y x getDate       function dayOfTheYear2 m d y       for any day   m is 1 to 12  d is 1 to 31  y is a 4-digit year var m d y  var M  31 28 31 30 31 30 31 31 30 31 30 31   if  y   400    0     y   4    0  amp  amp  y   100    0      M 1    var Y 0  for  var i 0 i lt m-1   i   Y  M i    return Y d

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This might be useful to those who need the day of the year as a string and have jQuery UI available    You can use jQuery UI Datepicker   day of year string     datepicker formatDate  o   new Date      Underneath it works the same way as some of the answers already mentioned   date ms - first date of year ms    ms per day    function getDayOfTheYearFromDate d        return Math round  new Date d getFullYear    d getMonth    d getDate    getTime    - new Date d getFullYear    0  0  getTime      86400000      day of year int   getDayOfTheYearFromDate new Date

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This is a simple way to find the current day in the year  and it should account for leap years without a problem   Javascript   Math round  new Date   setHours 23  - new Date new Date   getYear   1900  0  1  0  0  0   1000 60 60 24    Javascript in Google Apps Script   Math round  new Date   setHours 23  - new Date new Date   getYear    0  1  0  0  0   1000 60 60 24    The primary action of this code is to find the number of milliseconds that have elapsed in the current year and then convert this number into days  The number of milliseconds that have elapsed in the current year can be found by subtracting the number of milliseconds of the first second of the first day of the current year  which is obtained with new Date new Date   getYear   1900  0  1  0  0  0   Javascript  or new Date new Date   getYear    0  1  0  0  0   Google Apps Script   from the milliseconds of the 23rd hour of the current day  which was found with new Date   setHours 23   The purpose of setting the current date to the 23rd hour is to ensure that the day of year is rounded correctly by Math round     Once you have the number of milliseconds of the current year  then you can convert this time into days by dividing by 1000 to convert milliseconds to seconds  then dividing by 60 to convert seconds to minutes  then dividing by 60 to convert minutes to hours  and finally dividing by 24 to convert hours to days   Note  This post was edited to account for differences between JavaScript and JavaScript implemented in Google Apps Script  Also  more context was added for the answer

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Following OP s edit    x000D   x000D  var now   new Date    x000D  var start   new Date now getFullYear    0  0   x000D  var diff   now - start  x000D  var oneDay   1000   60   60   24  x000D  var day   Math floor diff   oneDay   x000D  console log  Day of year      day   x000D   x000D   x000D    Edit  The code above will fail when now is a date in between march 26th and October 29th andnow s time is before 1AM  eg 00 59 59   This is due to the code not taking daylight savings time into account  You should compensate for this     x000D   x000D  var now   new Date    x000D  var start   new Date now getFullYear    0  0   x000D  var diff    now - start      start getTimezoneOffset   - now getTimezoneOffset      60   1000   x000D  var oneDay   1000   60   60   24  x000D  var day   Math floor diff   oneDay   x000D  console log  Day of year      day   x000D   x000D   x000D

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If you don t want to re-invent the wheel  you can use the excellent date-fns  node js  library   var getDayOfYear   require  date-fns get day of year    var dayOfYear   getDayOfYear new Date 2017  1  1      1st february   gt  32

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USE THIS SCRIPT     var today   new Date    var first   new Date today getFullYear    0  1   var theDay   Math round   today - first    1000   60   60   24     5  0   alert  Today is the     theDay    theDay    1    st     theDay    2    nd     theDay    3    rd     th         day of the year

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x000D   x000D  const dayOfYear   date   gt    x000D      const myDate   new Date date   x000D      const year   myDate getFullYear    x000D      const firstJan   new Date year  0  1   x000D      const differenceInMillieSeconds   myDate - firstJan  x000D      return  differenceInMillieSeconds    1000   60   60   24    1   x000D     x000D   x000D  const result   dayOfYear  2019-2-01    x000D  console log result   x000D   x000D   x000D

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You can pass parameter as date number in setDate function   var targetDate   new Date    targetDate setDate 1       Now we can see the expected date as  Mon Jan 01 2018 01 43 24 console log targetDate    targetDate setDate 365       You can see  Mon Dec 31 2018 01 44 47 console log targetDate

[javascript] JavaScript calculate the day of the year (1 - 366)

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