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PHP determines whether a year is a leap year

(*-*)浩Original: 2019-09-27 15:31:167438browse

Determining leap years: (Recommended learning: PHP programming from entry to proficiency)

①. An ordinary year is a leap year if it is divisible by 4 but not divisible by 100. (For example, 2004 is a leap year, and 1900 is not a leap year)

②. A leap year is a century year that is divisible by 400 but not divisible by 3200. (For example, 2000 is a leap year, and 3200 is not a leap year)

Code:

<html>  
<head>
<meta charset="UTF-8" />  
<title>闰年判断</title>  
</head>  
<?php
$year = $_GET["year"];
if (isset($_GET["year"])) {
    if (is_numeric($year)) {
        if ($year % 100 == 0) { //判断世纪年
            if ($year % 400 == 0 && $year % 3200 != 0) {
                echo "世纪年" . $year . "是闰年!"; //世纪年里的闰年
            } else {
                echo "世纪年" . $year . "不是闰年!";
            }
        } else { //剩下的就是普通年了
            if ($year % 4 == 0 && $year % 100 != 0) {
                echo "普通年" . $year . "是闰年!"; //普通年里的闰年
            } else {
                echo "普通年" . $year . "不是闰年!";
            }
        }
    } else
        $msg = "请输入正确的格式";
}
?>  
<body>  
<form name=rn method=&#39;get&#39;>  
<b>请输入年：</b>  
<input type="text" name=year>  
<input type="submit" name=sub value="查询">  
<?php
echo $msg;
?>  
</form>    
</body>  
</html>

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