Calculate last day of month in JavaScript

2019-01-01 12:42发布

If you provide 0 as the dayValue in Date.setFullYear you get the last day of the previous month:

d = new Date(); d.setFullYear(2008, 11, 0); //  Sun Nov 30 2008

There is reference to this behaviour at mozilla. Is this a reliable cross-browser feature or should I look at alternative methods?

15条回答
若你有天会懂
2楼-- · 2019-01-01 13:16

I recently had to do something similar, this is what I came up with:

/**
* Returns a date set to the begining of the month
* 
* @param {Date} myDate 
* @returns {Date}
*/
function beginningOfMonth(myDate){    
  let date = new Date(myDate);
  date.setDate(1)
  date.setHours(0);
  date.setMinutes(0);
  date.setSeconds(0);   
  return date;     
}

/**
 * Returns a date set to the end of the month
 * 
 * @param {Date} myDate 
 * @returns {Date}
 */
function endOfMonth(myDate){
  let date = new Date(myDate);
  date.setMonth(date.getMonth() +1)
  date.setDate(0);
  date.setHours(23);
  date.setMinutes(59);
  date.setSeconds(59);
  return date;
}

Pass it in a date, and it will return a date set to either the beginning of the month, or the end of the month.

The begninngOfMonth function is fairly self-explanatory, but what's going in in the endOfMonth function is that I'm incrementing the month to the next month, and then using setDate(0) to roll back the day to the last day of the previous month which is a part of the setDate spec:

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Date/setDate https://www.w3schools.com/jsref/jsref_setdate.asp

I then set the hour/minutes/seconds to the end of the day, so that if you're using some kind of API that is expecting a date range you'll be able to capture the entirety of that last day. That part might go beyond what the original post is asking for but it could help someone else looking for a similar solution.

Edit: You can also go the extra mile and set milliseconds with setMilliseconds() if you want to be extra precise.

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忆尘夕之涩
3楼-- · 2019-01-01 13:21

In computer terms, new Date() and regular expression solutions are slow! If you want a super-fast (and super-cryptic) one-liner, try this one (assuming m is in Jan=1 format). I keep trying different code changes to get the best performance.

My current fastest version:

After looking at this related question Leap year check using bitwise operators (amazing speed) and discovering what the 25 & 15 magic number represented, I have come up with this optimized hybrid of answers:

function getDaysInMonth(m, y) {
    return m===2 ? y & 3 || !(y%25) && y & 15 ? 28 : 29 : 30 + (m+(m>>3)&1);
}

Given the bit-shifting this obviously assumes that your m & y parameters are both integers, as passing numbers as strings would result in weird results.

JSFiddle: http://jsfiddle.net/TrueBlueAussie/H89X3/22/

JSPerf results: http://jsperf.com/days-in-month-head-to-head/5

For some reason, (m+(m>>3)&1) is more efficient than (5546>>m&1) on almost all browsers.

The only real competition for speed is from @GitaarLab, so I have created a head-to-head JSPerf for us to test on: http://jsperf.com/days-in-month-head-to-head/5


It works based on my leap year answer here: javascript to find leap year this answer here Leap year check using bitwise operators (amazing speed) as well as the following binary logic.

A quick lesson in binary months:

If you interpret the index of the desired months (Jan = 1) in binary you will notice that months with 31 days either have bit 3 clear and bit 0 set, or bit 3 set and bit 0 clear.

Jan = 1  = 0001 : 31 days
Feb = 2  = 0010
Mar = 3  = 0011 : 31 days
Apr = 4  = 0100
May = 5  = 0101 : 31 days
Jun = 6  = 0110
Jul = 7  = 0111 : 31 days
Aug = 8  = 1000 : 31 days
Sep = 9  = 1001
Oct = 10 = 1010 : 31 days
Nov = 11 = 1011
Dec = 12 = 1100 : 31 days

That means you can shift the value 3 places with >> 3, XOR the bits with the original ^ m and see if the result is 1 or 0 in bit position 0 using & 1. Note: It turns out + is slightly faster than XOR (^) and (m >> 3) + m gives the same result in bit 0.

JSPerf results: http://jsperf.com/days-in-month-perf-test/6

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牵手、夕阳
4楼-- · 2019-01-01 13:21

This one works nicely:

Date.prototype.setToLastDateInMonth = function () {

    this.setDate(1);
    this.setMonth(this.getMonth() + 1);
    this.setDate(this.getDate() - 1);

    return this;
}
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像晚风撩人
5楼-- · 2019-01-01 13:23

A slight modification to solution provided by lebreeze:

function daysInMonth(iMonth, iYear)
{
    return new Date(iYear, iMonth, 0).getDate();
}
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墨雨无痕
6楼-- · 2019-01-01 13:23

This will give you current month first and last day.

If you need to change 'year' remove d.getFullYear() and set your year.

If you need to change 'month' remove d.getMonth() and set your year.

var d = new Date();
var days = ["Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"];
var fistDayOfMonth = days[(new Date(d.getFullYear(), d.getMonth(), 1).getDay())];
	var LastDayOfMonth = days[(new Date(d.getFullYear(), d.getMonth() + 1, 0).getDay())]; 
console.log("First Day :" + fistDayOfMonth); 
console.log("Last Day:" + LastDayOfMonth);
alert("First Day :" + fistDayOfMonth); 
alert("Last Day:" + LastDayOfMonth);

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梦该遗忘
7楼-- · 2019-01-01 13:24

My colleague stumbled upon the following which may be an easier solution

function daysInMonth(iMonth, iYear)
{
    return 32 - new Date(iYear, iMonth, 32).getDate();
}

stolen from http://snippets.dzone.com/posts/show/2099

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