Question

大家好，我需要php专家的建议。这是我的问题。用户有2个输入开始和结束这都是整数。这将能够识别奇数和偶数。我已经解决了奇怪甚至差不多完成的事情。主要问题是起始值1结束5. 1是奇数，应该在表格中显示奇数。但问题是它在偶数表中找到了。初始值是问题。其余的都很好。

这是我的代码

<?php
$firstnum = $_POST['first_input'];
$secondnum = $_POST['second_input'];
$counter = 0;
echo "<table border='1'>";
if ($firstnum < $secondnum) {
    echo "<tr>"; //first tr
    echo "<th>"; echo "Even numbers"; echo "</th>";
    echo "<th>"; echo "Odd numbers"; echo "</th>";
    echo "</tr>";
    for ($counter=$firstnum; $counter <= $secondnum ; $counter++) { 

        if ($counter % 2 == 0){
            echo "<tr>";
            echo "<td>";
            echo $counter; 
            echo "</td>";
        } else {
            echo "<td>";
            echo $counter; 
            echo "</td>";
            echo "</tr>";
        }
    }

} elseif ($firstnum > $secondnum) {
    # code...
    //first num is < second num
    echo "<tr>"; //first tr
    echo "<th>"; echo "Even numbers"; echo "</th>";
    echo "<th>"; echo "Odd numbers"; echo "</th>";
    echo "</tr>";
    for ($counter=$firstnum; $counter >= $secondnum ; $counter--) { 

        if ($counter % 2 == 0){
            echo "<tr>";
            echo "<td>";
            echo $counter; 
            echo "</td>";
        } else {
            echo "<td>";
            echo $counter; 
            echo "</td>";
            echo "</tr>";
        }
    }
}
echo "</table>";
?>

Answer 1

您的问题是，if / else阻止了您的html无效。

如果您有if

<tr>
    <td><td>

并在您的else中

   <td></td>
</tr>

这两个都需要完整的行/单元格标记

<tr>
    <td></td>
    <td></td>
</tr>

所以你的代码应该是

    if ($counter % 2 == 0){
        echo "<tr>";
        echo "<td>";
        echo $counter; 
        echo "</td>";
        echo "<td>"; 
        echo "</td>";
        echo "</tr>";
    } else {
        echo "<tr>";
        echo "<td>";
        echo "</td>";
        echo "<td>";
        echo $counter; 
        echo "</td>";
        echo "</tr>";
    }

表格的简单php脚本错误结果

1 个答案: