A PHP code example or code snippet to check leap year. This code is showing you how to create a simple Python program that checks whether a year is a leap year.
According to Wikipedia: a leap year is a year that has one additional day in order to adjust the calendar to the astronomical year.
The logic to check for leap year:
- if (year is not divisible by 4) then (it is a common year)
- else if (year is not divisible by 100) then (it is a leap year)
- else if (year is not divisible by 400) then (it is a common year)
- else (it is a leap year)
The years input that we will give & its expected output are:
- 2020 is a leap year
- 2019 is not a leap year
- 2018 is not a leap year
- 2000 is a leap year
- 1900 is not a leap year
PHP Code Example to Check for Leap Year
You can immediately run the following code in various IDEs that have PHP interpreters such as VS Code, Atom, or CMD.
The following PHP program code uses the Nested If and Comparison Operators.
<?php
echo "Input year: ";
$year = trim(fgets(STDIN));
if (($year % 4) == 0) {
if (($year % 100) == 0) {
if (($year % 400) == 0) {
echo "$year is a leap year";
} else {
echo "$year is not a leap year";
}
} else {
echo "$year is a leap year";
}
} else {
echo "$year is not a leap year";
}
Output
The code above will output the following result:
Input year:
2020 is a leap year
Input year: 2019
2019 is not a leap year
Input year: 2018
2019 is not a leap year
Input year: 2000
2000 is a leap year
Input year: 1900
1900 is not a leap year
Hopefully this PHP code snippet for checking leap years is useful.
