how to use Zend Route Regex to route a url to another one?

Question

i have a url like this: view_test.php?user=56

and i am trying to get it here:

    $router->addRoute(
        'test',
        new Zend_Controller_Router_Route(
            'test/view/:id',
            array(
                'module' => 'test',
                'controller' => 'index',
                'action' => 'view'
            )
        )
    );

basically from view_test.php?user=56 to /test/view/56

im not sure how to handle the ?.

any ideas?

thanks

David Weinraub · Accepted Answer · 2012-09-23T02:22:32.160

AFAIK, in a route, you can't reference the url's query-string content, just its request path. But you could do the following:

Add a route that maps the old url to a redirect action:

$router->addRoute('view-test-redirect', new Zend_Controller_Router_Route_Static(
    'view_test.php',
    array(
        'module' => 'test',
        'controller' => 'index',
        'action' => 'view-test-redirect',
    )
);

Also, add a route representing your "real" action:

$router->addRoute('view-test', new Zend_Controller_Router_Route(
    'view/test/:user',
    array(
        'module' => 'test',
        'controller' => 'index',
        'action' => 'view-test',
    )
);

The action name view-test-redirect corresponds to the action below:

public function viewTestRedirectAction()
{
    $user = (int) $this->_getParam('user', null);
    if (!$user){
        throw new \Exception('Missing user');
    }
    $this->_helper->redirector->goToRouteAndExit(array('user' => $user), 'view-test');
}

Then your view-test action can do things in the expected way:

public function viewTestController()
{
    $user = (int) $this->_getParam('user', null);
    if (!$user){
        throw new \Exception('Missing user');
    }
    // Read db, assign results to view, praise unicorns, etc.
}

Not tested, just brain-dumping to demonstrate the idea.

how to use Zend Route Regex to route a url to another one?

1 Answers1