php variables not storing

Question

I am trying to create a db class with php that takes the host ect as variable. I cant get the initialized values to stick and i am not sure why. When i initialize them at the top where i set them to public it works fine, but when i try to initialize them in the constructor it is not working.

    class Database {

    public $dbHost;
    public $dbUser;
    public $dbPass;
    public $dbName;

    public $db;

    public function __construct($Host, $User, $Pass, $Name){ 
        $dbHost = $Host;
        $dbUser = $User;
        $dbPass = $Pass;
        $dbName = $Name;
        $this->dbConnect();
    }

    public function dbConnect(){
        echo $dbPass;
        $this->db = new mysqli($this->dbHost, $this->dbUser, $this->dbPass, $this->dbName);

        /* check connection */
        if (mysqli_connect_errno()){
            printf("Connect failed: %s\n", mysqli_connect_error());
            exit();
        }else{
            //echo 'connection made';
        }
    }

andrewsi · Accepted Answer · 2012-10-23 18:04:43Z

6

You're not initializing them properly in the constructor; try:

$this->dbHost = $Host;

What you're currently doing is initializing a local variable called $dbHost, whose scope is just the constructor function itself.

answered Oct 23, 2012 at 18:04

andrewsi

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Comments

Musa · Accepted Answer · 2012-10-23 18:05:30Z

2

You have to use $this to access instance variables inside a class e.g. $this->dbHost = $Host;

answered Oct 23, 2012 at 18:05

Musa

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Comments

Jacek Sokolowski · Accepted Answer · 2012-10-23 18:05:52Z

2

Change this:

public function __construct($Host, $User, $Pass, $Name){ 
        $dbHost = $Host;
        $dbUser = $User;
        $dbPass = $Pass;
        $dbName = $Name;
        $this->dbConnect();
    }

to this:

public function __construct($Host, $User, $Pass, $Name){ 
        $this->dbHost = $Host;
        $this->dbUser = $User;
        $this->dbPass = $Pass;
        $this->dbName = $Name;
        $this->dbConnect();
    }

answered Oct 23, 2012 at 18:05

Jacek Sokolowski

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Comments

user1726343 · Accepted Answer · 2012-10-23 18:06:30Z

1

Try this:

public function __construct($Host, $User, $Pass, $Name){ 
        $this->dbHost = $Host;
        $this->dbUser = $User;
        $this->dbPass = $Pass;
        $this->dbName = $Name;
        $this->dbConnect();
    }

answered Oct 23, 2012 at 18:06

user1726343

Comments

JDischler · Accepted Answer · 2012-10-23 18:05:52Z

0

How about using this->

public function __construct($Host, $User, $Pass, $Name){ 
    $this->dbHost = $Host;
    $this->dbUser = $User;
    $this->dbPass = $Pass;
    $this->dbName = $Name;
    $this->dbConnect();
}

answered Oct 23, 2012 at 18:05

JDischler

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