CODEWITHSUNDEEP
doctrinesymfony
JERC
JERC

Reputation: 1624

execute php app/console doctrine:schema:update from controller symfony 2

I want to execute this command

php app/console doctrine:schema:update

from the controles without use exec php function,

Any comment will be use full to me.

Thanks!!!

Upvotes: 11

Views: 10879

Answers (2)

JERC
JERC

Reputation: 1624

Thanks to every one,

I used this code

            $kernel = $this->get('kernel');
            $application = new \Symfony\Bundle\FrameworkBundle\Console\Application($kernel);
            $application->setAutoExit(false);
            //Create de Schema 
            $options = array('command' => 'doctrine:schema:update',"--force" => true);
            $application->run(new \Symfony\Component\Console\Input\ArrayInput($options));
            //Loading Fixtures
            $options = array('command' => 'doctrine:fixtures:load',"--append" => true);
            $application->run(new \Symfony\Component\Console\Input\ArrayInput($options));

and in this link, there are some information maybe will be usefull for others

Thanks!!!

Upvotes: 13

Sgoettschkes
Sgoettschkes

Reputation: 13214

If you are looking here, on the bottom there is an example of how to execute a command from within symfony code.

Please also mind the note on the end saying that it might not be a good idea to use a command within your code.

As said, the following code should be used with care. I wouldn't use it for the reasons statet in the symfony doc, but it works.

When using the following code within your controller, you are able to execute a command:

$kernel = $this->get('kernel');
$application = new \Symfony\Bundle\FrameworkBundle\Console\Application($kernel);
$application->setAutoExit(false);

$options = array('command' => 'list');
$application->run(new \Symfony\Component\Console\Input\ArrayInput($options));

If you need the output, you have to either use an existing class implementing OutputInterface or create your own depending on your needs.

Upvotes: 15

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