Hi,

i have the Problem, that the Tests are missing if i install the version 2.1 via composer. If i look in your repository in github, it seems composer.json are the reason for that:

"archive": { "exclude": ["/tests"] }

Problem here is the Code in vendor/php-mock/php-mock/autoload.php where you create class alias for a class who does not exists in the project. I cant really use your package with this behaviour.

The namespace of MockNamespaceTest class did not comply with psr-4 autoloading standard. Leaving the class with incorrect namespace would cause it to not be autoloaded in Composer v2.0.

When mocking system-functions that expect variables to be passed by reference, this hack allows to do so. This might f.i. help mocking sort-functions or exec or passthru

Hi all, I'm new to PHP unit testing so please advise if I'm dong anything wrong. This is the scenario:

In our main code, there is a function (say b) that calls another function (say a). I want to test b() without having to test a() so want to make a mock of a. However in my test call when I can b() it still calls the original a, not my mock. Here is my code:

In my test file:

<?php
namespace AA\BB\CC;
use PHPUnit\Framework\TestCase;
use phpmock\MockBuilder;

class XYZ extends TestCase {
        private $a_mock;

        public function testb() {
                $a_builder    = new MockBuilder();
                $a_builder->setNamespace(__NAMESPACE__)
                        ->setName("a")
                        ->setFunction(
                            function ($str) {
                                return $str;
                            }
                        );

                $this->a_mock = $a_builder->build();
                $this->a_mock->enable();
                $result = b("Hello");
                        // NOTE: If I call a() here, then the mock version is called
                        // but if b() is called, it calls the other a() defined below
                $this->a_mock->disable();
        }
}

In our code base (I don't see any namespaces defined here):

function a($str) {
        // do something
}

function b($str) {
        return a($str);
}

Hi Markus,

Thank you for accepting my previous PR!

I would find useful if MockEnvironment extends Mock. This would write the following:

class MyTest extends \PHPUnit_Framework_TestCase
{
    public function setUp()
    {
        $builder = new SleepEnvironmentBuilder();
        $builder->setNamespace('My\\Namespace')
            ->setTimestamp(strtotime('2030-01-01 00:00:00'));
        $mock = $builder->build();
        $mock->enable();
        $this->getTestResultObject()->addListener(new MockDisabler($mock));
        // ...

Indeed, MockDisabler's constructor requires a Mock instance.

Regards, Geoffroy

The source article referenced 7 years ago when PHP5.3 was new used the standard non-bracketed namespace declaration. If you use the bracketed one here you actually unlock global namespace mocks as well.

Specifically: \function_call() becomes mockable.

Fatal error: Declaration of phpmock\phpunit\MockObjectProxy::__phpunit_verify() must be compatible with PHPUnit_Framework_MockObject_MockObject::__phpunit_verify(bool $unsetInvocationMocker = true) in vendor/php-mock/php-mock-phpunit/classes/MockObjectProxy.php on line 17

It looks like Mockery is not working with the new PHP 7 Return Types.

If I have an method with a return type:

<?php

namespace Game\Repository;

use Game\Entity\UserInterface;

Interface UserRepositoryInterface
{
    public function createNewUser($username, $password, $email):UserInterface;
}

And I mock it: $userDummy = Mockery::mock('Game\Entity\UserInterface');

I get this error:

PHP Fatal error:  Declaration of 
Mockery_2_Game_Repository_UserRepositoryInterface::createNewUser($username, $password, $email) 
must be compatible with 
Game\Repository\UserRepositoryInterface::createNewUser($username, $password, $email): Game\Entity\UserInterface 
in ***\vendor\mockery\mockery\library\Mockery\Loader\EvalLoader.php(16) : eval()'d code on line 25                

If I take the return type off, Mockery works again. Mockery's php version says ">=5.5" so I was hoping it would work with 7 :)

Hi, guy!

I want to replace my static method by using setName. But I think It can't be static method, right? If YES, I want to have some advice. there is sample code below. I'll wait for answer.

thanks!

    // foo\bar::doSomething() is replaced with 'Yepppp!'
    $builder = new \phpmock\MockBuilder();
    $builder->setNamespace('foo')
            ->setName('doSomething')
            ->setFunction(function () {
                return 'Yepppp!!';
            });
    $mock = $builder->build();
    $mock->enable();

This is a handy library, and i'm using it to replace a old test case, which originally with namespace hack like:

namespace Test;
class ATest {}

namespace App;
function extension_loaded() {}

The test purpose is to simulate PHP extension load status, and throw Exception when failed.

For using php-mock, i create a trait like this:

trait ExtensionLoadedMockTrait
{
    /** @type bool */
    public static $extensionLoaded = true;


    /**
     * @param   string  $namespace
     * @return  Mock
     */
    public function buildExtensionLoadedMock($namespace)
    {
        $function = 'extension_loaded';
        $mockContainer = MockContainer::getInstance();
        $mock = $mockContainer->getMock($namespace, $function);

        if (is_null($mock)) {
            $mock = (new MockBuilder())
                ->setNamespace($namespace)
                ->setName($function)
                ->setFunction(function($ext) {
                    return self::$extensionLoaded &&
                        \extension_loaded($ext);
                })
                ->build();

            $mock->define();

            $mockContainer->registerMock($namespace, $function, $mock);
        }

        return $mock;
    }
}

The source code of MockContainer is omit. In test case, use

$this->buildExtensionLoadedMock('Foo\Bar')->enable();

to enable or disable mock.

This test works fine when run by PHPUnit with single file given, and with directory given, but fails when run PHPUnit without any parameter:

// Ok
phpunit path/to/test/case.php
// Ok
phpunit path/to/tests/
// Fail
phpunit

Why ?

As you mentioned about Bug #68541, the mock need to define before native PHP functions. So I make a special test case name like 'Lib\Aaa\Aaa.php', and I'm sure it will run before all other test cases, but the mock mechanism still not work, the mocked function are not triggered at all.

Run single file are successful, i think the code is correct, but failed when batch run confuse me, do you have any tips ? Thanks.

PHP: 5.5.9 PHPUnit: 4.5.0 Os: Ubuntu 14.04 Trusty

Hi,

I hoped to mock die() function (or exit), seems not possible. I got that :

namespace cool;

use phpmock\\generator\\MockFunctionGenerator;

function exit()
{
    $arguments = [];

    $variadics = \\array_slice(\\func_get_args(), \\count($arguments));
    $arguments = \\array_merge($arguments, $variadics);

    return MockFunctionGenerator::call(
        \'exit\',
        \'cool\\exit\',
        $arguments
    );
}

PHP Parse error: syntax error, unexpected 'exit' (T_EXIT), expecting '(' in \vendor\php-mock\php-mock\classes\generator\MockFunctionGenerator.php(67) : eval()'d code on line 5