Check if year is leap year in javascript [duplicat

Question

This question already has an answer here:

• javascript to find leap year 7 answers

 function leapYear(year){
    var result; 
    year = parseInt(document.getElementById("isYear").value);
    if (years/400){
      result = true
    }
    else if(years/100){
      result = false
    }
    else if(years/4){
      result= true
    }
    else{
      result= false
    }
    return result
 }

This is what I have so far (the entry is on a from thus stored in "isYear"), I basically followed this here, so using what I already have, how can I check if the entry is a leap year based on these conditions(note I may have done it wrong when implementing the pseudocode, please correct me if I have) Edit: Note this needs to use an integer not a date function

Answer 1

The function checks if February has 29 days. If it does, then we have a leap year.

function isLeap(year) {
  return new Date(year, 1, 29).getDate() === 29;
}

ES6

let isLeap = (year) => new Date(year, 1, 29).getDate() === 29;

isLeap(1004) // true
isLeap(1001) // false

Answer 2

If you're doing this in an Node.js app, you can use the leap-year package:

npm install --save leap-year

Then from your app, use the following code to verify whether the provided year or date object is a leap year:

var leapYear = require('leap-year');

leapYear(2014);
//=> false 

leapYear(2016);
//=> true 

Using a library like this has the advantage that you don't have to deal with the dirty details of getting all of the special cases right, since the library takes care of that.

Answer 3

A faster solution is provided by Kevin P. Rice here:https://stackoverflow.com/a/11595914/5535820 So here's the code:

function leapYear(year)
{
    return (year & 3) == 0 && ((year % 25) != 0 || (year & 15) == 0);
}

Answer 4

You can try using JavaScript's Date Object

new Date(year,month).getFullYear()%4==0

This will return true or false.

Answer 5

function leapYear(year)
{
  return ((year % 4 == 0) && (year % 100 != 0)) || (year % 400 == 0);
}

Answer 6

My Code Is Very Easy To Understand

var year = 2015;
var LeapYear = year % 4;

if (LeapYear==0) {
    alert("This is Leap Year");
} else {
    alert("This is not leap year");
}