從mysql更改爲mysqli後出現錯誤

Question

嗨im試圖從mysql更改爲mysqli和ive使用了一個編輯器，它將所有更改並即時通訊更改所有文件後出現以下錯誤 致命錯誤：調用成員函數query（）非對象/home/matureco/public_html/config/db_connect.php第7行

繼承人的一塊編碼爲

$conn  = new mysqli(DBHOST,DBUSER,DBPASS) or die($mysqli->error); 

$row  = mysqli_fetch_array($mysqli->query("select * from settings where id = '1' ")); 
$site_name = trim(stripslashes($row['site_name'])); 
$email  = trim($row['email']); 
$keyword = trim(stripslashes($row['keyword'])); 
$description= trim(stripslashes($row['description'])); 
$logo  = trim($row['logo']); 
$copyright = trim(stripslashes($row['copyright'])); 

$favicon = trim($row['favicon']); 
$paypal_email = trim($row['paypal_email']); 

誰能告訴我是什麼香港專業教育學院錯過

非常感謝ty jan x

Answer 1

$mysqli不是您的代碼中的對象。 $conn是您聲明的對象。

您的代碼應該表明這一點：

$row  = mysqli_fetch_array($conn->query("select * from settings where id = '1' ")); 

而且，你缺少的對象聲明的參數設置數據庫本身：

正確的語法：

$obj = new mysqli("my_host", "my_user", "my_password", "my_db"); 

Answer 2

$mysqli->query("select * from settings where id = '1' "));應be

$conn->query("select * from settings where id = '1' ")); 

因爲您要使用oop方法。你應該改變你的代碼。 這是我的建議

$result  = $conn->query("select * from settings where id = '1' "); 
$row = $result->fetch_object() 
$site_name = trim(stripslashes($row->site_name)); 
$email  = trim($row->email); 
$keyword = trim(stripslashes($row->keyword)); 
$description= trim(stripslashes($row->description)); 
$logo  = trim($row->logo); 
$copyright = trim(stripslashes($row->copyright)); 

$favicon = trim($row->favicon); 
$paypal_email = trim($row->paypal_email);