如何控制PHP中闭包的命名空间？

Question

我得到了

传递给Tests\Unit\BaseUnitTestCase::assertMockedMethodRuns()的参数1必须是测试\单元\闭包的实例，即给定的闭包实例

将闭包传递给我的assertMockedMethodRuns方法时

BaseUnitTestCase

namespace Tests\Unit;

use Tests\TestCase;

class BaseUnitTestCase extends TestCase
{
    public function assertMockedMethodRuns(Closure $code)
    {
        try {
            $code($this);
        } catch (\Mockery\Exception\InvalidCountException $e) {
            $this->assertTrue(False, "Mocked method did not run");
        }
    }
}

路由测试(我称之为assertMockedMethodRuns)

namespace Tests\Unit;

use App\Http\Controllers\PostController;

class RoutesTest extends BaseUnitTestCase
{
    public function testGetPostList()
    {
        $mockPostController = \Mockery::mock('PostController');
        $mockPostController->shouldReceive('index')->once();
        $this->app->instance('PostController', $mockPostController);
        $this->assertMockedMethodRuns(function() { 
            $this->call('GET','/post');
        });
    }
}

TestCase ( BaseTestCase从这里继承)

namespace Tests;

use Illuminate\Foundation\Testing\TestCase as BaseTestCase;
use Illuminate\Foundation\Testing\DatabaseMigrations;
use Illuminate\Foundation\Testing\RefreshDatabase;

abstract class TestCase extends BaseTestCase
{
    use CreatesApplication, DatabaseMigrations;
}

传递给assertMockedMethodRuns的闭包似乎没有正确的命名空间。我该怎么解决这个问题？

Answer 1

您将丢失并反向斜杠以告诉PHP Closure类将在全局命名空间中搜索：

public function assertMockedMethodRuns(\Closure $code)