警告mysqli_query（）需要至少2个参数[关闭]

I am Using Following Code. But I get this warning message. I don't know why this happens.

mysqli_query() expects at least 2 parameters....

Config:

    class org_type
{
    private $conn = '';
    function __construct()
    {
        global $con;
        $this->conn = $con;
    }

    public function create(){
        $this->create_org_type();
    }

    $sql = "CREATE TABLE IF NOT EXISTS `org_type`
       (
        `orgTypeId` int NOT NULL AUTO_INCREMENT,
        PRIMARY KEY(orgTypeId),
        `orgTypeName` text,
        `status` varchar(150),
        `create_date` INT(11),
        `desp` TEXT DEFAULT ''
        )";
    mysqli_query($this->conn,$sql);

}*/

public function insert_orgType($orgName,$sts,$editVal){
   $now = time();
    mysqli_query($this->conn, "INSERT INTO `org_type`(orgTypeName,status,create_date,desp) VALUES ('".$orgName."','".$sts."',".$now.",'".$editVal."')") or die (mysqli_error($this->conn));
    $insId = mysqli_insert_id($this->conn);
    return $insId;

}

    public function select_orgType($id=0){
        $str = '';
        $arrResult = array();


        if($id != 0){
            $str .= " AND orgTypeId = $id ";        
        }            
        $sql = "SELECT * FROM org_type WHERE 1 $str order by orgTypeId ASC";
        $result = $this->conn->query($sql);
if ($result->num_rows > 0) { 
            while($row = $result->fetch_assoc()) { 
            $arrResult[] = $row;
        }
        return $arrResult;
        }
    }    
}

mysqli_query() requires connection variable to be passed. Update your query as following:

mysqli_query($this->conn,"INSERT INTO `org_type`(orgTypeName,status,create_date,desp) VALUES ('".$orgName."','".$sts."',".$now.",'".$editVal."')") or die (mysqli_error());

You need to pass the connection object as a parameter for following function

1st : mysqli_query() expects two parameter

First one is connection object

Second one is query

 mysqli_query($this->conn,query);

2nd : mysqli_error()

you need to pass connection object to mysqli_error($this->conn) function

3rd : mysqli_insert_id()

you need to pass connection object to mysqli_insert_id($this->conn) function

Update 1: you need to include that connection file

function __construct()
{
     global $con;
     include('connection.php');
     $this->conn = $con;
}