从mysql更改为mysqli后出现错误

Question

嗨im试图从mysql更改为mysqli和ive使用了一个编辑器，它将所有更改并即时通讯更改所有文件后出现以下错误 致命错误：调用成员函数query（）非对象/home/matureco/public_html/config/db_connect.php第7行

继承人的一块编码为

$conn  = new mysqli(DBHOST,DBUSER,DBPASS) or die($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']); 

谁能告诉我是什么香港专业教育学院错过

非常感谢ty jan x

Answer 1

$mysqli不是您的代码中的对象。 $conn是您声明的对象。

您的代码应该表明这一点：

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

而且，你缺少的对象声明的参数设置数据库本身：

正确的语法：

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

Answer 2

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

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

因为您要使用oop方法。你应该改变你的代码。 这是我的建议

$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);