OO PHP undefined variables

Question

I'm trying my hand at OO PHP however when I try the code I get errors saying that dbhost, dbuser,dbpass and dbname are undefined. Netbeans also gives me a warning saying that they might be uninitialized. Removing the static keyword gives me an error saying 'Unexpected "$dbhost" '. Does anyone know what I'm doing wrong?

<?php
class DatabaseManager {

private static $dbhost = 'localhost';
private static $dbuser = 'root';
private static $dbpass = '';
private static $dbname = 'app_db';

public static function getConnection(){
    $dbconn;
    try {
    $dbconn = new PDO('mysql:host='.$dbhost,'dbname='.$dbname,
    $dbuser, $dbpass);
    } catch (PDOException $e) {
        echo "Could not connect to database";
        echo $e;
        exit;
    }
    return $dbconn;

}

}
?>

Unrelated to the problem: You don't have to declare variables in PHP, so there is no point in doing $dbconn; at the beginning of your method (this just has no effect at all). — Niko
– Niko, Commented Aug 11, 2012 at 14:26

Ray · Accepted Answer · 2012-08-11 14:07:42Z

3

You've declared your variable static. Reference them like this in php 5.2 or higher:

$dbconn = new PDO('mysql:host='.self::$dbhost,'dbname='.self::$dbname,
                   self::$dbuser, self::$dbpass);

In PHP 5.3 or higher if you set them from private to protected you can also use:

$dbconn = new PDO('mysql:host='.static::$dbhost,'dbname='.static::$dbname,
                   static::$dbuser, static::$dbpass);

They both act similarly, but if you extend the class, the static keyword allows for late static binding.

answered Aug 11, 2012 at 14:07

Ray

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5 Comments

Niko Over a year ago

Since the member variables are declared private, I wouldn't recommend using static. That may otherwise lead to some problems with subclasses.

Ray Over a year ago

@Niko see my comment above the static example, it indicates he should switch from private to protected to use static.

Jack Hurt Over a year ago

Another quick question, does PHP handle the scope of variables differently than java? In my code above $dbconn; is unused and I can return the dbconn variable that is initialized inside the try

Ray Over a year ago

@JackHurt mostly it's the same. Since php doesn't have packages it's simpler using the protected keyword. You can declare $dbconn to be a static protected property as well outside of the method and use it without having to return it.

Niko Over a year ago

@JackHurt Yes, there is a difference. Java has a block scope, so that a variable is only defined inside the closest pair of curly brackets. In PHP it's a function scope, meaning that a variable is always defined in the whole function (as you may have already guessed from the name ^^).

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