Question

我是laravel的新手，想对laravel 5.6中的控制器进行单元测试这是我的控制器。

class NewOrdersController extends Controller
{
    protected $newOrders;

    public function __construct(NewOrderRepositoryInterface $newOrders)
    {
        $this->newOrders = $newOrders;
    }

   public function orderList(){
         $orders = $this->newOrders->getNewOrders();
         if(count($orders) == 0){
                return 0;
         }
         else{
                return 1;
         }
   }
}

我的测试看起来像这样。如何编写控制器的单元测试？

class NewOrderTest extends TestCase
{
    public function tearDown()
    {
        Mockery::close();
    }

    /** @test */
    public function testMock(){
        $neworders = Mockery::mock('NewOrderRepositoryInterface');
        $neworders->shouldReceive('getNewOrders')
                        ->once()
                        ->andReturn(0);

        $orders = new newOrder($this->neworders);
        $this->assertEquals(0,$orders->orderList());
    }
}

Answer 1

模拟是要替换您已经在使用的模拟，您绝对不能断言或测试一种模拟。通常，您只需要制作mockX()方法，该方法就可以将模拟绑定到您要替换的类。另外，您在接口上有一个有趣的绑定，只需将类路径用作绑定即可，而且更容易，您可以从::class派生它。将模拟设置为实例后，该模拟将自动加载到类中。

public function mockOrderRepository(){
    $neworders = Mockery::mock(NewOrderRepositoryInterface::class);
    $neworders->shouldReceive('getNewOrders')
        ->once()
        ->andReturn(0);

    // bind the mocked class to the dependency injected version.
    $this->app->instance(NewOrderRepositoryInterface::class, $neworders);
}

我认为它是一个JsonApi，通常是:)否则模拟您的存储库并调用端点，您可以使用assertJson来断言数据

/** @test */
public function testOrderList()
{
    $this->mockOrderRepository;

    $response = $this->json('GET', 'v1/your/api/endpoint');

    $response->assertStatus(200);

    $response->assertJson(1);
}

如何在Laravel中编写控制器的单元测试？

1 个答案: