PHP Program to Check Leap Year

Posted in PHP Tutorial

Updated on Oct 31, 2024

By Mari Selvan

👁️ 157 - Views

⏳ 4 mins

💬 1 Comment

Photo Credit to CodeToFun

🙋 Introduction

In the realm of programming, determining whether a given year is a leap year is a common requirement. A leap year is a year that is evenly divisible by 4, except for end-of-century years, which must be divisible by 400 to be a leap year. Checking for leap years involves implementing specific rules to validate the year.

In this tutorial, we will explore a PHP program designed to check whether a given year is a leap year.

📄 Example

Let's delve into the PHP code that accomplishes this task.

isLeapYear.php

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<?php
// Function to check if a year is a leap year
function isLeapYear($year)
{
  // Leap year conditions
  if (($year % 4 == 0 && $year % 100 != 0) || ($year % 400 == 0)) return true; // True, it's a leap year
  else return false; // False, it's not a leap year
  
}

// Driver program
// Replace this value with the year you want to check
$year = 2024;

// Call the function to check if the year is a leap year
if (isLeapYear($year)) echo "$year is a leap year.\n";
else echo "$year is not a leap year.\n";

?>

💻 Testing the Program

To test the program with different years, modify the value of $year in the code.

Output

2024 is a leap year.

Run the script to check if the specified year is a leap year.

🧠 How the Program Works

The program defines a function isLeapYear that takes an integer $year as input and returns true if the year is a leap year, and false otherwise.
The function checks leap year conditions: the year must be divisible by 4 and not divisible by 100, or it must be divisible by 400.
Inside the main program, replace the value of $year with the desired year you want to check.
The program calls the isLeapYear function and prints the result using echo.

📏 Between the Given Range

Let's dive into the php code that checks for leap years in the specified range.

isLeapYear.php

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<?php
// Function to check if a year is a leap year
function isLeapYear($year)
{
  return ($year % 4 == 0 && $year % 100 != 0) || ($year % 400 == 0);
}

// Range of years from 2024 to 2050
$startYear = 2024;
$endYear = 2050;

// Display leap years in the specified range
echo "Leap Years in the range $startYear to $endYear:\n";

for ($year = $startYear;$year <= $endYear;$year++)
{
  if (isLeapYear($year))
  {
    echo $year . " ";
  }
}

?>

💻 Testing the Program

The provided code is already set to check for leap years in the range from 2024 to 2050.

Output

Leap years in the range 2024 to 2050:
2024 2028 2032 2036 2040 2044 2048

You can run the script as is or modify the range as needed.

🧠 How the Program Works

The program defines a function isLeapYear that checks if a given year is a leap year using the leap year rule.
It sets the range of years from 2024 to 2050 using variables $startYear and $endYear.
The program then iterates through each year in the range, using a for loop.
For each year, it checks if it's a leap year using the isLeapYear function and prints the leap years.

🧐 Understanding the Concept of Leap Year

Before delving into the code, let's understand the concept of leap years. Leap years are years that are evenly divisible by 4, except for end-of-century years, which must be divisible by 400 to be a leap year.