如何根据提交的表单ID显示查询结果？(php)

<form method="POST" action="process_form.php">
  <label for="form_id">表单ID：</label>
  <input type="text" name="form_id" id="form_id">
  <input type="submit" value="查询">
</form>

<?php
$form_id = $_POST['form_id'];
?>

<?php
// 连接到数据库
$servername = "数据库服务器地址";
$username = "数据库用户名";
$password = "数据库密码";
$dbname = "数据库名";

$conn = new mysqli($servername, $username, $password, $dbname);

// 检查连接是否成功
if ($conn->connect_error) {
    die("连接数据库失败: " . $conn->connect_error);
}

// 根据表单ID执行查询
$sql = "SELECT result FROM forms WHERE form_id = '$form_id'";
$result = $conn->query($sql);

// 处理查询结果
if ($result->num_rows > 0) {
    // 输出查询结果
    while($row = $result->fetch_assoc()) {
        echo "查询结果：" . $row["result"];
    }
} else {
    echo "未找到相关结果";
}

// 关闭数据库连接
$conn->close();
?>