无法通过php上传文本值到mysql

Question

每次我提交我的网站的评论表格时，它只会创建一个新的ID.But其他值是空的，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(); 

你正在尝试使用不存在的变量。您需要检索张贴的形式从$_POST数组中的值：

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

编辑：请注意，您在每次请求运行此代码，即使形式不贴。您将要添加一个检查为：

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

其实，你可能想要把具有与在if语句来里面的数据库中保存的数据做的一切，让你不开数据库连接或准备一份声明，当你不打算使用它时。