Fatal error: Call to a member function query() PHP CLASS

Question

I\'m trying to write this class to connect and query my database, but I got this error:

Fatal error: Call to a member function query() on null in C:

Answer 1

Your method:

public function db_connect() {    

            $connection = mysqli_connect(self::$db_host, self::$db_user, self::$db_pass, self::$db_name) or die("Error " . mysqli_error($connection)); 
            echo "Conexión realizada". "
";

    }  

does not retun the $connection back from the method so the rest method calls fail do this:

public function db_connect() {    

            $connection = mysqli_connect(self::$db_host, self::$db_user, self::$db_pass, self::$db_name) or die("Error " . mysqli_error($connection)); 
            echo "Conexión realizada". "
";
           return $connection; // return the $connection object
    }  

As mentioned, your code is not efficient since for every query performed the code (re-)connects to the DB. THis is unnecessarily expensive/inefficient.

There are many approaches to solve this.

1. Connect to the DB on the instantiation of the DB class

e.g

class Db{

    private static $db_host = "localhost";
    private static $db_user = "root";
    private static $db_pass = "";
    private static $db_name = "sivi";

    public $connection;

    public function __construct()
   {
      $this->connection = $this->db_connect();
   }  

    public function db_connect() {    

            $connection = mysqli_connect(self::$db_host, self::$db_user, self::$db_pass, self::$db_name) or die("Error " . mysqli_error($connection)); 
            echo "Conexión realizada". "
";
            return $connection;
    }  
}

2. Lazy connection, i.e connect only on first executed query

e.g

class Db{

    private static $db_host = "localhost";
    private static $db_user = "root";
    private static $db_pass = "";
    private static $db_name = "sivi";
    public $connection = null;
    public function __construct()
    {
    }  
    public function db_connect() {    
        $connection = mysqli_connect(self::$db_host, self::$db_user, self::$db_pass, self::$db_name) or die("Error " . mysqli_error($connection)); 
        echo "Conexión realizada". "
";
        return $connection;
    }  
    public function db_query($query){
        if ( null ==== $this->connection ) $this->connection = $this->db_connect();
        var_dump($query);
        $result = $this->connection->query($query);
        while($row = mysqli_fetch_array($result)) { 
            echo $row["COD_PRE"] . "
";
        }
    }
}