訪問資源文件中的Symfony

Question

目前我有目錄結構，如： `

src 
-App 
--MyBundle 
---Service 
----DefaultService 
tests 
-AppMyBundle 
--files 
---data.csv 
--Service 
---DefaultServiceTest 

I'm writting test for a file reader and I added data.csv file for demo file for a test. In my test I need to get a path of that data.csv file. But I couldn't find any way of getting it. I have tried accessing it via file_locator service like:

$locator = $this->container->get('file_locator'); 
$path = $locator->locate('@AppMyBundle/Tests/files/data.csv'); 

and

$locator = $this->container->get('file_locator'); 
$path = $locator->locate('@AppMyBundle/files/data. 

With no luck. namespaces if necessary of my DefaultService is應用程序\ MyBundle \服務and for DefaultServiceTest is應用程序\ MyBundle \測試\ Service`

Answer 1

您是從應用程序/束系統之外，從而嘗試這種代碼，請：

$path = $this->get('kernel')->getRootDir() . '/../tests/AppMyBundle/files/data.csv'; 

Answer 2

我用$client->getContainer()->getParameter('kernel.root_dir')這樣的：

$client = static::createClient(); 

$schemaPath = sprintf(
    '%s/../tests/MyBundle/Resources/rss2.xsd', 
    $this->_client->getContainer()->getParameter('kernel.root_dir') 
);