$dbh is not a recognized variable when it is set?

Question

I'm getting the following message:

Notice: Undefined variable: dbh in /var/www/PDO/Functions/PDOFunctions.php on line 12 Fatal error: Call to a member function prepare() on a non-object in /var/www/PDO/Functions/PDOFunctions.php on line 12

$dbh = new PDO('mysql:host=localhost;dbname=pdo', 'root', 'xxxxxxxxxxx');
global  $dbh;


function PDOFetch($Var)
{
    $sth = $dbh->prepare("$Var"); //Error Line
    $sth->execute();
    $result = $sth->fetchAll();
    return $result; 
}

function PDONumb ($Var)
{
    $Query = $dbh->prepare("{$Var}");
    $Execute->execute();
    $count = $Execute->rowCount();
    return $count;
}

What is the problem with my code?

The problem is you don't know how global works. You should have googled this and read the documentation before asking a question on Stack Overflow... — meagar

gview gview · Accepted Answer · 2012-12-01T01:45:54

Using global variables is bad practice. For something as easy as this, you could rewrite your code as a simple class. In doing so, you also get the added benefit of easily being able to create and use multiple database handles.

class Db 
{
    private $dbh = null;

    public function __construct()
    {
        $this->dbh = new PDO('mysql:host=localhost;dbname=pdo', 'root', 'xxxxxxxxxxx');
    }

    public function PDOFetch($Var)
    {
        $sth = $this->dbh->prepare("$Var"); //Error Line
        $sth->execute();
        $result = $sth->fetchAll();
        return $result; 
    }

    public function PDONumb ($Var)
    {
        $sth = $this->dbh->prepare("{$Var}");
        $sth->execute();
        $count = $sth->rowCount();
        return $count;
    }
    // Other methods here
}

Then it's:

$dbc1 = new Db();
$dbc2 = new Db();  // Hey I have 2 connections now, cool
$result1 = $dbc1->PDOFetch(..);
$result2 = $dbc2->PDOFetch(..);

Note, that your PDONumb was broken and wouldn't work, so I threw in a fix for that as well.

$dbh is not a recognized variable when it is set?

3 Answers