I need to create a function that would return the next 50 leap years beginning from the next leap year after the input year in R. The problem is that I need to save it as a vector. x <- function(year){ year1 <- year + 205 for(i in year:year1){ if((i
package leapYear; import java.util.Scanner; import javax.swing.JOptionPane; public class leapYear { public static void main (String[] args){ String yearInput = JOptionPane.showInputDialog("Enter the year here: "); Scanner inputScanner = new Scan
I want to calculate the number of days between 2 dates. Common problem. For exemple : var d0 = new Date("2016-02-27"); var d1 = new Date("2017-08-25"); Many people recommend using epoch difference : var res = (d1 - d0) / 1000 / 60 / 60
I'm trying to compare a measure as of today through the same day and month for the prior 4 years (e.g. through June 6 of 2016, 2015, 2014, etc.). For each year, I decided to count the number of days since the beginning of the year, and sum my values
I have some trouble to calculate the weekday of a given date next year. For example: Tuesday, 19. April 2016 is the given date. Now I would calculate: TUESDAY, 18. April 2017. It is very important that the weekdays are always the same. My problem are
I'm trying to write codes with PHP to enable leap year friendly select option tags. For example, you pick a year, then it checks if it is a leap year or not, then it shows a pull down menu (select option tags) of days preceded by another select optio
I'm working with C# and I'm trying to find whether the given date and month is valid for a leap year. This is my code: static void Main(string[] args) { Console.WriteLine("The following program is to find whether the Date and Month is Valid for an LE
avHi again, thank you all for the help in my previous problem. However now i have encountered another problem int main () { int inputSeconds,seconds, minutes, hours , days ,years ; int remainingSeconds ; int current_year = 1970; int current_month = 1
The program is supposed to return what day of the week it is for the entered date. One of the dates that doesn't work is 01012000. Nothing is returned at all. But on some other leap years the first day of March can be calculated. Also sometimes seemi
This could be simple but i just can't get around it at the moment practice question. I'm just trying to print a few integers together so that output would be 2004 is a leap year 2013 is not a leap year public class Ex1partA { public static void main(
I am doing a task in which it is required to find the last day of the year by given current date. Like if its 06/Jan/2013 is given to it and it will automatically return the end day of the year. like 31/Dec/2013 I am using many ways like adding the r
I want to get the number of days in the month which the user specifies. I am using this it works for most months except for Feb and leap year. It shows 28 days and not 29. Can you solve this? begin declare @year int declare @month int select @year =
When I need to insert date into the database I need to check 2 things: If this month has 31 days in case day 31 was selected If February is selected, I need to check if it's a leap year in case the user selected 29 Currently I'm checking this using a
While subtracting a year form current date, if current day is February 28 2013 then below statement returns February 28 2012, but correct result should be February 29 2012 as 2012 is a leap year. How can this scenario be handled. SELECT DATEADD(year,
Im using the following function to calculate the number of leap days between two years: static int CountLeapDays(int startYear, int endYear) { int Days = 0; while (true) { if ((startYear % 4 == 0 && startYear % 100 != 0) || startYear % 400 == 0) D
I have date when someone is born, and I need to calculate with JavaScript or jQuery or so, how many years, days since birth date, until now. So result can be like 20 years, 89 days. I need to have same results as Wikipedia does with their function of
Someone on JSPerf dropped an amazingly fast implementation for checking leap years of the ISO calendar (link: Odd bit manipulations): function isLeapYear(year) { return !(year & 3 || year & 15 && !(year % 25)); } Using Node.js, I quickly c
When i have a leap year in my database (ex.: 29th Feb 2012). The EntityFunctions.CreateDateTime functions throws System.Data.SqlClient.SqlException: Conversion failed when converting date and/or time from character string. My Code is as follows in my
I have a program where the user enters a date and then compares it to another date to see which one comes first. How would I go about writing a code where the user inputs Feb 29 and the program returns Feb 28 instead (since there is no leap year)? Ex
I have a HTML page with 3 dropdowns for the month, day and year and I was wondering if there was a way to populate the month drop down properly depending on the month and year. I haven't done this before on the client side, but it looks like a lot of
Is there a method in any native Java class to calculate how many days were/will be in a specific year? As in, was it a Leap year (366 days) or a normal year (365 days)? Or do I need to write it myself? I'm calculating the number of days between two d
$(.dateselboxes) .change( function(){ var y; y=$("#year").val(); var m; m=$("#month").val(); var d; // check leap year var leapYear; if(y%4==0) { if(y%100==0) { if(y%400==0) {leapYear=true;} else {leapYear=false;} } else {leapYear=true
I made a program using C to find whether the entered year is a leap year or not. But unfortunately its not working well. It says a year is leap and the preceding year is not leap. #include<stdio.h> #include<conio.h> int yearr(int year); void m
I have been tasked to read in some data from some weird old system. The system contains many dates but they are all oddly formatted. They are integer numbers ranging from approximately 55,000 to 80,000. I know two dates for certain: 58,112 equals Feb
In order to find leap years, why must the year be indivisible by 100 and divisible by 400? I understand why it must be divisible by 4. Please explain the algorithm.The length of a year is (more or less) 365.242196 days. So we have to subtract, more o
