PHP SQL从select options html中获取值

Question

我想从我的下拉菜单中获取所选的选项，但它会在任何情况下返回我的表的最后一个'id'。请帮帮我:)。

<?php
    $data = "SELECT * FROM Users";
    $reasult = $db->query($data);
    echo "<form name='filter' method='get'>
    <p><label>Felhasználók: </label></p>
    <select name='filter'>
    <option value='' selected='selected'></option>
    ";
        while($record = mysqli_fetch_array($reasult))
        {
            $selected_option=$record['ID'];
            if ($record['admin']==1)
            {
                $rang='admin';
            }
            else
            {
                $rang='user';
            }
            echo "<option value=" . $record['ID'] . ">" . $record['username'] . "</option>";
        }
        echo "</select></form>";
        echo $selected_option . "<br>";
?>

Answer 1

将echo $selected_option . "<br>"; 移到while循环的内。

while($record = mysqli_fetch_array($reasult))
        {
            $selected_option=$record['ID'];
            if ($record['admin']==1)
            {
                $rang='admin';
            }
            else
            {
                $rang='user';
            }
            echo "<option value=" . $record['ID'] . ">" . $record['username'] . "</option>";
            echo $selected_option . "<br>";
        }
        echo "</select></form>";