PHP綁定參數$ mysqli-> prepare（）語句不起作用

Question

嗨，我不太確定我寫的是什麼問題。朋友有相同的代碼，但我的工作不會。希望能得到幫助。

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; 
} 

感謝

Answer 1

您有$stmt->fetch()不讀取的結果。儘管您已將結果列綁定到$postTitle，但除非您從語句結果集中獲取一行，否則沒有值可用。

// 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; 
    } 
}