How do I parse a yaml file from within a service in Symfony2

18,991

Solution 1

First I'll explain why I implemented my solution for you to decide if this case is right for you.

I needed a way to easily load custom .yml files in my bundle (for lots of bundles) so adding a separate line to app/config.yml for every file seemed like a lot of hassle for every setup.

Also I wanted most of the configs to be already loaded by default so end-user wouldn't even need to worry about configuring most of the time, especially not checking that every config file is setup correctly.

If this seems like a similar case for you, read on. If not, just use Kris solution, is a good one too!


Back when I encountered a need for this feature, Symfony2 didnt't provide a simple way to achieve this, so here how I solved it:

First I created a local YamlFileLoader class which was basically a dumbed down Symfony2 one:

<?php

namespace Acme\DemoBundle\Loader;

use Symfony\Component\Yaml\Yaml;
use Symfony\Component\Config\Loader\FileLoader;

/**
 * YamlFileLoader loads Yaml routing files.
 */
class YamlFileLoader extends FileLoader
{
    /**
     * Loads a Yaml file.
     *
     * @param string $file A Yaml file path
     *
     * @return array
     *
     * @throws \InvalidArgumentException When config can't be parsed
     */
    public function load($file, $type = null)
    {
        $path = $this->locator->locate($file);

        $config = Yaml::parse($path);

        // empty file
        if (null === $config) {
            $config = array();
        }

        // not an array
        if (!is_array($config)) {
            throw new \InvalidArgumentException(sprintf('The file "%s" must contain a YAML array.', $file));
        }

        return $config;
    }

    /**
     * Returns true if this class supports the given resource.
     *
     * @param mixed  $resource A resource
     * @param string $type     The resource type
     *
     * @return Boolean True if this class supports the given resource, false otherwise
     *
     * @api
     */
    public function supports($resource, $type = null)
    {
        return is_string($resource) && 'yml' === pathinfo($resource, PATHINFO_EXTENSION) && (!$type || 'yaml' === $type);
    }
}

Then I updated DIC Extension for my bundle (it's usually generated automatically if you let Symfony2 create full bundle architecture, if not just create a DependencyInjection/<Vendor&BundleName>Extension.php file in your bundle directory with following content:

<?php

namespace Acme\DemoBundle\DependencyInjection;

use Symfony\Component\DependencyInjection\ContainerBuilder;
use Symfony\Component\Config\FileLocator;
use Symfony\Component\HttpKernel\DependencyInjection\Extension;
use Symfony\Component\DependencyInjection\Loader;

use Acme\DemoBundle\Loader\YamlFileLoader;

/**
 * This is the class that loads and manages your bundle configuration
 *
 * To learn more see {@link http://symfony.com/doc/current/cookbook/bundles/extension.html}
 */
class AcmeDemoExtension extends Extension
{
    /**
     * {@inheritDoc}
     */
    public function load(array $configs, ContainerBuilder $container)
    {
        $configuration = new Configuration();
        $config = $this->processConfiguration($configuration, $configs);

        $loader = new Loader\XmlFileLoader($container, new FileLocator(__DIR__.'/../Resources/config'));

        $loader->load('services.xml');

        // until here everything is default config (for your DIC services)

        $ymlLoader = new YamlFileLoader(new FileLocator(__DIR__.'/../Resources/config'));
        $container->setParameter('param_name', $ymlLoader->load('yaml_file_name'))); // load yml file contents as an array
    }
}

And now you can access/pass your yaml config as simple service parameter (i.e. %param_name% for services.yml)

Solution 2

I solved it this way:

Services.yml

#/path/to/app/src/Bundle/Resources/config/services.yml

parameters:
    example.class: Path\To\Bundle\Service\Class
    example.yaml_config_file: "%kernel.root_dir%/../src/Path/To/Bundle/Resources/config/config.yml"

services:
    example_service:
        class: %example.class%
        arguments: [%example.yaml_config_file% ]

Service class

# /path/to/app/src/Bundle/Service/Example.php

<?php

namespace Bundle\Service;

use \Symfony\Component\Yaml\Yaml;

class Example
{
    private $parsed_yaml_file;

    public function __construct($yaml_config_file)
    {
       $this->parsed_yaml_file = Yaml::parse($yaml_config_file);
    }
}

Solution 3

You can use the kernel.root_dir parameter:

parameters:
    yaml.config_file: "%kernel.root_dir%/../src/Path/To/MyBundle/Resources/config/config.yml"
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18,991
Pete Mitchell
Author by

Pete Mitchell

Web developer (primarily PHP/Javascript) living in Montreal, QC.

Updated on July 17, 2022

Comments

  • Pete Mitchell
    Pete Mitchell almost 2 years

    I want to get an array from a yaml file inside one of my services, and I am a little confused of how to inject the file to use in my services.yml.

    # /path/to/app/src/Bundle/Resources/config/services.yml
    parameters:
        do_something: Bundle\DoSomething
        yaml.parser.class: Symfony\Component\Yaml\Parser
        yaml.config_file: "/Resources/config/config.yml" # what do I put here to win!
    
    services:
        yaml_parser:
            class: %yaml.parser.class%
    
        do_parsing:
            class: %do_something%
            arguments: [ @yaml_parser, %yaml.config_file% ]
    

    In my service I have

    # /path/to/app/src/Bundle/DoSomething.php
    
    <?php
    
    namespace Bundle;
    
    use \Symfony\Component\Yaml\Parser;
    
    class DoSemething
    {
        protected $parser;
        protected $parsed_yaml_file;
    
        public function __construct(Parser $parser, $file_path)
        {
           $this->parsed_yaml_file = $parser->parse(file_get_contents(__DIR__ . $file_path));
        }
    
        public function useParsedFile()
        {
            foreach($parsed_yaml_file as $k => $v)
            {
                // ...  etc etc 
            }
        }
    }
    

    This may be the completely wrong approach, if I should be doing something else please let me know!

  • Pete Mitchell
    Pete Mitchell over 12 years
    Thanks for the in depth answer, very helpful.