从类中发出查询/准备语句[关闭]

Question

I have a class I am working with:

class SlayWork
{
     public function AllowDB ()
     {
          $MySQLi = new mysqli(DBAHost, DBAUsername, DBAPassword, DBADB);
     }
}

How would I communicate with my database from another public function from within this class? example:

public function GetTest ()
{
   $Var = $MySQL->prepare("SELECT * FROM test");
   $Var->execute();
   $Var->bind_result($ID, $Uname);
}

adding

public $MySQLi  = new mysqli(DBAHost, DBAUsername, DBAPassword, DBADB); 

without the public function returns:

Parse error: syntax error, unexpected 'new' (T_NEW) in C:\xampp\htdocs\Error\ClassWork.php on line 16

adding global with the public function returns:

Parse error: syntax error, unexpected '=', expecting ',' or ';' in C:\xampp\htdocs\Error\ClassWork.php on line 18

Answer 1

Create an empty property like private $db_conn;. Then create a class construct and inside it create the connection like this:

$this->db_conn = new mysqli(DBAHost, DBAUsername, DBAPassword, DBADB);

Now every time you instantiate your class you are connected to the database and all methods you call will be able to communicate with your db.

Example:

class.php

class Database {
    private $db_conn;

    public function __construct() {
        $this->db_conn = new mysqli(DBAHost, DBAUsername, DBAPassword, DBADB);
    }

    public function GetTest () {
       $Var = $this->db_conn->prepare("SELECT * FROM test");
       $Var->execute();
       $Var->bind_result($ID, $Uname);
    }
}