服務器上的錯誤！需要另一雙眼睛在簡單的php的mysqli

Question

我有PHP代碼從MySQL數據庫中選擇用戶詳細信息

$mysql = new mysqli(DB_SERVER, DB_USER, DB_PASSWORD, DB_NAME) or die('There was a problem connecting to the database'); 
$stmt = $mysql->prepare('SELECT personalinfo.userid, personalinfo.firstname, personalinfo.lastname FROM personalinfo INNER JOIN applications ON personalinfo.userid = applications.userid WHERE applications.jobref = ?'); 
$stmt->bind_param('i',$jr); 
$stmt->execute(); 
$stmt->store_result(); 
$stmt->bind_result($userID,$firstName,$lastName); 
$count = $stmt->num_rows(); 

我已經localy測試此代碼在它的工作原理以下罰款，但是當我試圖測試上直播服務器，我不斷收到以下錯誤 「致命錯誤：調用一個成員函數bind_param（）一個非對象在C：\路徑\爲\目錄下的」提前偷看

THX，

Aaron

任何

Answer 1

那麼，根據對mysqli::prepare()的文件，它將返回FALSE如果有一個錯誤，這是試圖調用$stmt->bind_param()因爲$stmt不是時候爲什麼你得到一個「無對象」錯誤這是一個對象。

因此，無論是爲了您目前的錯誤還是將來的任何錯誤，請務必在創建語句時檢查錯誤並採取相應措施。

$mysql = new mysqli(DB_SERVER, DB_USER, DB_PASSWORD, DB_NAME) or die('There was a problem connecting to the database'); 
$stmt = $mysql->prepare('SELECT personalinfo.userid, personalinfo.firstname, personalinfo.lastname FROM personalinfo INNER JOIN applications ON personalinfo.userid = applications.userid WHERE applications.jobref = ?'); 

if ($stmt) 
{ 
    $stmt->bind_param('i',$jr); 
    $stmt->execute(); 
    $stmt->store_result(); 
    $stmt->bind_result($userID,$firstName,$lastName); 
    $count = $stmt->num_rows(); 
} else { 
    // something went wrong 
    // perhaps display the error? 
    echo $mysql->error; 
}