警告：mysqli_query（）预计参数1是mysqli的，给定的对象

Question

我遇到了这个错误的问题，我用的代码如下：

<?php 
$cus_name = isset($_POST['client_name'])?$_POST['client_name']:''; 
$description = isset($_POST['desc'])?$_POST['desc']:''; 
$amount = isset($_POST['amnt'])?$_POST['amnt']:''; 

$query = "INSERT INTO ".$RECORD_TABLE."(cus_name,description,amount) VALUES ('$cus_name','$description','$amount') " ; 
$result = mysqli_query($dbObj,$query); 

if(!$result) 
{ 
    echo "error while inserting"; 
} 


?> 

这是我使用的代码将数据插入数据库。连接代码，我下面给：

$DB_HOST = 'localhost'; 
$DB_USER = 'root'; 
$DB_PWD = ''; 
$DB_NAME = 'hotel_booking'; 

$BOOKING_TABLE = 'hotel_booking'; 
$PRICING_TABLE = 'hotel_pricing'; 
$SETTING_TABLE = 'hotel_setting'; 
$RECORD_TABLE = 'hotel_record'; 
require_once('Database.class.php'); 

global $dbObj; 
$dbObj = new Database($DB_HOST,$DB_USER,$DB_PWD,$DB_NAME,1,0); 

Answer 1

我觉得你的问题是用于连接 试试这个代码

$conn = mysqli_connect($DB_HOST,$DB_USER,$DB_PWD,$DB_NAME); 

，改变

$query = "INSERT INTO ".$RECORD_TABLE."(cus_name,description,amount) VALUES ('".$cus_name."','".$description."','".$amount."') " ; 
$result = $conn->query($query); 

Answer 2

现在你$dbObj是Database类的对象。如果你想用mysqli_*，那么$dbObj应该是这样的。

$dbObj = mysqli_connect("host", "user", "password","database name"); 

有关更多信息，请参阅here。

Answer 3

检查类是否返回新的mysqli

class Database() { 
    function __constructor(){ 
     $obj=mysqli_connect($DB_HOST,$DB_USER,$DB_PWD,$DB_NAME); 
     return $obj; 
    } 
}