PHP MySQL : Unable to Connect to Database

Question

i am trying to connect to database in php , Database name is photo_gallery following is my code

Database.php

    define("DB_SERVER", "localhost"); 
    define("SERVER_NAME", "root"); 
    define("DB_PASS" ,"");
    define("DB_NAME", "photo_gallery");

    class MySQLDatabase{

        function __construct(){
            $this->open_connection();
        }

        private $connection ; 
        public function open_connection (){
        $this->connection = mysqli_connect(DB_SERVER, DB_NAME, DB_PASS);
            if(!$this->connection){
                die("Connection failed".mysqli_error());
        }
        else {
            $db_select=  mysqli_select_db($this->connection,DB_NAME);
            if(!$db_select){
                echo 'Database selection failed '.mysqli_error($this->connection);
            }
        }
    }
        public  function query($sql){
            $query = mysqli_query($this->connection,$sql); 
            $this->confirm_query($query);
            return $query;
        }       

        private function confirm_query($result_set){
            $query= mysqli_query($this->connection,$result_set); 
            if(!$query){
                    die('Database query failed '.mysqli_error());
            } 
        }

        public function close_connection(){
            if(isset($this->connection)){
                mysqli_close($this->connection);
                unset($this->connection);
            }
        }       
    }

now when i run i get following error

Database selection failed Access denied for user: ' @' @ localhost ' . Base ' photo_gallery

i have confirmed username / password and db details but still unable to resolve . Please guide me about it

Answer 1

You haven't supplied a username - your mysqli_connect(DB_SERVER, DB_NAME, DB_PASS) line doesn't define a user, so you're trying to connect to your database with no user.

Try mysqli_connect(DB_SERVER, SERVER_NAME, DB_NAME, DB_PASS) instead.

Answer 2

second parameter of the connect method of mysqli-connect is the USERNAME . use it like this

$this->connection = mysqli_connect(DB_SERVER, DB_USERNAME, DB_PASS, DB_NAME );

http://php.net/manual/en/mysqli.construct.php

Answer 3

Your problem is that you chose bad names for your constants..

SERVER_NAME is not a good name for the database username..

As a consecuence of this, you used it wrong.. I suggest to change the constant to DB_USER and use it properly in the mysqli_connect() function..

Remember to use names that really represent what they do/refer..

Answer 4

This is the syntax:

   $conn= mysqli_connect("myhost","myuser","mypassw","mydb");

Here is the code:

Database.php

    define("DB_SERVER", "localhost"); 
    define("SERVER_NAME", "root"); 
    define("DB_PASS" ,"");
    define("DB_NAME", "photo_gallery");

    class MySQLDatabase{

        function __construct(){
            $this->open_connection();
        }

        private $connection ; 
        public function open_connection (){
        $this->connection = mysqli_connect(DB_SERVER, SERVER_NAME, DB_PASS, DB_NAME);
            if(!$this->connection){
                die("Connection failed".mysqli_error());
        }
        else {
            $db_select=  mysqli_select_db($this->connection,DB_NAME);
            if(!$db_select){
                echo 'Database selection failed '.mysqli_error($this->connection);
            }
        }
    }
        public  function query($sql){
            $query = mysqli_query($this->connection,$sql); 
            $this->confirm_query($query);
            return $query;
        }       

        private function confirm_query($result_set){
            $query= mysqli_query($this->connection,$result_set); 
            if(!$query){
                    die('Database query failed '.mysqli_error());
            } 
        }

        public function close_connection(){
            if(isset($this->connection)){
                mysqli_close($this->connection);
                unset($this->connection);
            }
        }       
    }