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php – 闰年和非闰年进入单独的表

作者：互联网

这段代码将显示从1991年到2100年的闰年和非闰年,我试图在闰年和非闰年制作一个表,但我失败了.

如何以表格格式或网格系统中引入它？这是为了学术研究.

 <!DOCTYPE html>
    <html>
    <head>
        <title>Leap Year</title>
        <script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/js/bootstrap.min.js"></script>
        <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.1/jquery.min.js"></script>
        <link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css">
    </head>
    <body>
        <?php  
        function isLeap($year) {
        return ((($year % 4) == 0) && ((($year % 100) != 0)));
        }

            for($year=1991; $year<=2100; $year++)  
            {  
                If (isLeap($year))  
                {   
                    $leap="$year : LEAP YEAR <br/>";
                    //echo "<div class='col-sm-12'>" . $leap . "</div>";
                    echo $leap;
                }  
                else  
                {  
                    $nonLeap="$year : Not leap year <br/>";  
                    //echo "<div class='col-sm-6'>" . $nonLeap ."</div>";       
                    echo $nonLeap;           
                }  
            }
        ?>
    </body>
    </html>

解决方法:

你的isLeap函数是错误的.你也可以参考这个post.

function isLeap($year) {
    return ((($year % 4) == 0) && ((($year % 100) != 0) || (($year % 400) == 0)));
}

for($year=1991; $year<=2100; $year++)  
{  
    if (isLeap($year))  
    {   
        $leaps[] = $year;
    }  
    else  
    {  
        $nonLeaps[] = $year;
    }
}

echo '<table><tr>';
foreach($leaps as $y)
{
  echo '<td>' . $y . '</td>';
}
echo '</tr></table>';

标签：php,css,bootstrap-modal,html,leap-year
来源： https://codeday.me/bug/20190710/1427917.html