[FIXED] Symfony4 how to call a file in public folder?

Issue

I need to display a yaml file as an array but i but I can't display it.

I have create a service, and my controller call this service.

In my service i try to call my yaml like this :

$value = Yaml::parseFile('public\assets\organizations.yaml');
return $value;

But that return me that error :

File "public\assets\organizations.yaml" does not exist.

Solution

You are specifying the file with a relative path. This will not be resolved relative to the file it is in but relative to the current working directory you are in when executing the script. Since this depends on various factors it will always be troublesome.

Thus you should always use absolute paths. In symfony you can get the base path of your project via the kernel.project_dir configuration parameter. Your code does not provide enough context to understand how or where you are using it. But if it is inside a controller extending AbstractController you can use getParameter():

$projectDir = $this->getParameter('kernel.project_dir');
$absolutePath = $projectDir . '/public/assets/organizations.yml';

$value = Yaml::parseFile($absolutePath);
return $value;

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Also note that using \ as the directory separator won't work under Linux/Unix-like systems where / is used as directory separator! Since / will work as directory separator under Windows, too, it is easiest to use it for cross-OS compatibility. Alternatively use the DIRECTORY_SEPARATOR constant.

Answered By - acran