CODEWITHSUNDEEP
phpunit-testingintegration-testing
Manoj Dhiman
Manoj Dhiman

Reputation: 5166

Php unit testing pass config file from command

I am implementing some Integration tests by using unit testing. I am using the setUp() method to set a config file. My config file consists of some Api keys API secret etc. This is working good.

public function setUp()
{
    $this->_pipeline = new Pipline($this);
    $this->_pipeline->setConfig(\Api\Models\ConfigModel::createFromFile(__DIR__ . '/test/integration/assets/config'));
    $users = $this->_pipeline->users->listUsers();
    $this->_userId = $users->result->data[0]->user_id;

    parent::setUp(); 
}

This method is working good in this way.

Problem But I want to pass the config file from command. Currently i am using this command

php ./vendor/phpunit/phpunit/phpunit -c ./test/integration/phpunit.xml

Is there a way i can pass this from command something like this

php ./vendor/phpunit/phpunit/phpunit -c ./test/integration/phpunit.xml -c2 /test/integration/assets/config

Thanks in advance.

Upvotes: 0

Views: 1061

Answers (1)

Nick
Nick

Reputation: 147206

You could just pass the name as an argument to php:

php ./vendor/phpunit/phpunit/phpunit -c ./test/integration/phpunit.xml -- /test/integration/assets/config

You can access the arguments via $argc and $argv; $argv[0] is the name of the script, so your config file name will be in $argv[1].

e.g.

E:\>php -r "echo \"$argc\n\"; print_r($argv);" -- "hello world!"
2
Array
(
    [0] => -
    [1] => hello world!
)

Upvotes: 1

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