Reputation: 31
How can I start a Symfony2 script using the php shell? I can't run the Controller file directly by using the command:
php FController.php
The path of the controller is
domain.com/web/app_dev.php/fcontroller
Do I have to use the Symfony2 console to run this script?
Upvotes: 3
Views: 4283
Reputation: 629
As has already been said you need to create a console command. Create a directory called 'Command' in one of your bundles (the bundle needs to be registered in AppKernel.php. Then create a class in this directory, it will be automatically found by symfony when you run app/console.
Here is a quick example:
<?php
namespace Acme\FooBundle\Command;
use Symfony\Bundle\FrameworkBundle\Command\Command,
Symfony\Component\Console\Input\InputOption,
Symfony\Component\Console\Input\InputInterface,
Symfony\Component\Console\Output\OutputInterface;
class BarCommand extends Command
{
protected function configure()
{
$this
->setName('foo:bar-cmd')
->setDescription('Test command')
->addOption('baz', null, InputOption::VALUE_NONE, 'Test option');
;
}
/**
* Execute the command
* The environment option is automatically handled.
*/
protected function execute(InputInterface $input, OutputInterface $output)
{
$output->writeln('Test command');
}
}
You can then run the command with:
$> app/console foo:bar-cmd
And pass in options like:
$> app/console foo:bar-cmd --baz
Upvotes: 6
Reputation: 530
A console command is the correct thing to use - its purpose is to be your gateway into your application from the command line. If you use the console cookbook chapter, setting up a console command is pretty easy:
http://symfony.com/doc/2.0/cookbook/console.html
You could technically create a PHP script that looks much like your front controller, except that you've modified the Request object to "fake" the web path (e.g. /foo/bar) to execute your controller, but a console command is really the right just for this.
Good luck!
Upvotes: 3