How do I generate a friendly URL in Symfony PHP?

Question

I always tend to forget these built-in symfony functions for making links.

Answer 1

This advice is for symfony 1.0. It probably will work for later versions.

Within your sfAction class:

string genUrl($parameters = array(), $absolute = false)

eg. $this->getController()->genUrl('yourmodule/youraction?key=value&key2=value', true);

In a template:

This will generate a normal link.

string link_to($name, $internal_uri, $options = array());

eg. link_to('My link name', 'yourmodule/youraction?key=value&key2=value');

Answer 2

If your goal is to have user-friendly URLs throughout your application, use the following approach:

1) Create a routing rule for your module/action in the application's routing.yml file. The following example is a routing rule for an action that shows the most recent questions in an application, defaulting to page 1 (using a pager):

recent_questions:
   url:    questions/recent/:page
   param:  { module: questions, action: recent, page: 1 }

2) Once the routing rule is set, use the url_for() helper in your template to format outgoing URLs.

<a href="<?php echo url_for('questions/recent?page=1') ?>">Recent Questions</a>

In this example, the following URL will be constructed: http://myapp/questions/recent/1.html.

3) Incoming URLs (requests) will be analyzed by the routing system, and if a pattern match is found in the routing rule configuration, the named wildcards (ie. the :/page portion of the URL) will become request parameters.

You can also use the link_to() helper to output a URL without using the HTML <a> tag.

Answer 3

In addition, if you actually want a query string with that url, you use this:

link_to('My link name', 'yourmodule/youraction?key=value&key2=value',array('query_string'=>'page=2'));

Otherwise, it's going to try to route it as part of the url and likely break your action.