无法通过php将文本值上传到mysql

Question

每次我提交我的网站表单进行评论时，它只会创建一个新的ID。但是所有其他值都是空的，我该怎么办？我是否需要为名称，电子邮件和内容创建一些属性？

<?php
error_reporting(E_ALL);ini_set('display_errors',1); 

$servername = "localhost";
$username = "seamaszhou";
$password = "123456";
$dbname = "guest";




$conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, 
$password);
// set the PDO error mode to exception
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

// prepare sql and bind parameters
$stmt = $conn->prepare("INSERT INTO guest 
(guestName,guestEmail,guestContent) 
VALUES (:guestName, :guestEmail, :guestContent)");

$stmt->bindParam(':guestName', $guestName);
$stmt->bindParam(':guestEmail', $guestEmail);
$stmt->bindParam(':guestContent', $guestContent);

 // insert a row
 $guestName = "$guestName";
$guestContent = "$guestContent";
$guestEmail = "$guestEmail";
$stmt->execute();



?>

HTML代码：

Answer 1

这是错误的：

// insert a row
$guestName = "$guestName";
$guestContent = "$guestContent";
$guestEmail = "$guestEmail";
$stmt->execute();

您尝试使用的变量不存在。您需要从$_POST数组中检索表单发布的值：

$guestName = $_POST['guestName'];
$guestContent = $_POST['guestContent'];
$guestEmail = $_POST['guestEmail'];
$stmt->execute();

编辑：请注意，即使表单未发布，您也会在每个请求上运行此代码。您需要添加一个检查：

if (isset($_POST['guestName']) && isset($_POST['guestContent']) && isset($_POST['guestEmail'])) {
    $guestName = $_POST['guestName'];
    $guestContent = $_POST['guestContent'];
    $guestEmail = $_POST['guestEmail'];
    $stmt->execute();
}

实际上，您可能希望将与保存数据库中的数据有关的所有内容放在if - 语句中，这样您就不会打开数据库连接或准备语句。 ＃39;不会使用它。