This PHP program will print the output whether the entered year is a leap year or not.
Three steps are there to find a leap year:
1. year % 4 == 0 -> evenly divisible by 4
2. and year % 100 != 0 -> should not be evenly divisible by 100
3. or year % 400 == 0 -> evenly divisible by 400
if 1st and 2nd is true or 3rd is true then it is a leap year.
Here is the entire PHP program to find leap year:
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<html> <body> <h2>PHP Script to find Leap year or not - Script by <a href="http://www.tutorialsmade.com/">Tutorialsmade.com</a></h2> <form action="" method="post"> <input type="text" name="year" /> <input type="submit" /> </form> </body> </html> <?php if( $_POST ) { //get the year $year = $_POST[ 'year' ]; //check if entered value is a number if(!is_numeric($year)) { echo "Strings not allowed, Input should be a number"; return; } //multiple conditions to check the leap year if( (0 == $year % 4) and (0 != $year % 100) or (0 == $year % 400) ) { echo "$year is a leap year"; } else { echo "$year is not a leap year"; } } ?> |
Read the comments in the program to understand the flow.
Here is the demo of this program:
hi can u explain this code
1. year % 4 == 0 -> evenly divisible by 4
2. and year % 100 != 0 -> should not be evenly divisible by 100
3. or year % 400 == 0 -> evenly divisible by 400
i did not get