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Home/ Questions/Q 7501735
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Editorial Team
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Editorial Team
Asked: May 29, 20262026-05-29T20:38:20+00:00 2026-05-29T20:38:20+00:00

I have the following code in IndexController.php: $sql = ‘SELECT * FROM ?’; $stmt

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I have the following code in “IndexController.php”:

$sql = 'SELECT * FROM ?';
$stmt = new Zend_Db_Statement_Mysqli($db, $sql);
$this->view->projects = $stmt->execute(array('projects'));

… which is just used to retrieve all project objects from the database and pass them to the view.
However when I run this code I get the following error:

Notice: Undefined variable: db in C:\wamp\www\PROJECTS_Zend\projectManager\application\controllers\IndexController.php on line 19

Fatal error: Call to a member function quoteIdentifier() on a non-object in C:\wamp\www\PROJECTS_Zend\projectManager\library\Zend\library\Zend\Db\Statement.php on line 181

I am not sure what variable db is, or what it should be, but if you have any information on this I would be greatful if you could enlighten me.

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  1. Editorial Team
    Editorial Team
    2026-05-29T20:38:22+00:00Added an answer on May 29, 2026 at 8:38 pm

    Well this way you are ruining the whole M(odel) concept of the MVC.

    Anyways $db is an instance of Zend_Db. I would advice you to read http://framework.zend.com/manual/en/zend.db.adapter.html.

    Although this might lead to awful code:

    $db = new Zend_Db_Adapter_Pdo_Mysql(array(
        'host'     => '127.0.0.1',
        'username' => 'webuser',
        'password' => 'xxxxxxxx',
        'dbname'   => 'test'
    ));
    
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