PHP绑定参数$ mysqli-> prepare（）语句不起作用

Question

嗨，我不太确定我写的是什么问题。朋友有相同的代码，但我的工作不会。希望能得到帮助。

if (isset($_GET['postID'])) 
{ 
    $postID = $_GET['postID']; 
    $stmt = $mysqli->prepare("SELECT postTitle FROM Posts WHERE postID = ?"); 
    $stmt->bind_param('i', $postID); 
    $stmt->execute(); 
    $stmt->bind_result($postTitle); 
    echo $postTitle; 
} 

感谢

Answer 1

您有$stmt->fetch()不读取的结果。尽管您已将结果列绑定到$postTitle，但除非您从语句结果集中获取一行，否则没有值可用。

// First, don't forget to establish your connection 
$mysqli = new MySQLi($host, $user, $pass, $dbname); 

if (isset($_GET['postID'])) 
{ 
    $postID = $_GET['postID']; 
    $stmt = $mysqli->prepare("SELECT postTitle FROM Posts WHERE postID = ?"); 
    $stmt->bind_param('i', $postID); 
    $stmt->execute(); 
    $stmt->bind_result($postTitle); 

    // Use a while loop if multiple rows are expected, 
    // Otherwise, a single call to $stmt->fetch() will do. 
    while ($stmt->fetch()) { 
    echo $postTitle; 
    } 
}