Unit testing 如何正确地获取单元测试中的错误以及禁用csrf检查？

我正在尝试在控制器中测试我的

post

方法。方法定义类似于：

    public function store(Request $request)
    {
        $article = new Article;

        $article->id = $request->input('article_id');
        $article->title = $request->input('title');
        $article->body = $request->input('body');
        return response(["success"], 200);
    }

我创建了一个测试，它只存储数据并检查响应是否为200。 还请告诉我如何使此测试更好地进行新测试。但是我得到了404错误，我不知道是什么错误。如何显示错误我需要配置哪些设置？ 测试：

public function test_post_new_article(){
        $article = factory(Article::class)->make();
        $this->call('POST', 'article', [
            '_token' => csrf_token(),
            'article_id' => 6,
            'title'=>"hey",
            'body' => "this is a body"
        ])->assertStatus(200);
    }

There was 1 failure:

1) Tests\Unit\ExampleTest::test_post_new_article
Expected status code 200 but received 404.
Failed asserting that false is true.

phpunit错误：

public function test_post_new_article(){
        $article = factory(Article::class)->make();
        $this->call('POST', 'article', [
            '_token' => csrf_token(),
            'article_id' => 6,
            'title'=>"hey",
            'body' => "this is a body"
        ])->assertStatus(200);
    }

There was 1 failure:

1) Tests\Unit\ExampleTest::test_post_new_article
Expected status code 200 but received 404.
Failed asserting that false is true.

---

我假设您在

routes/api.php

中定义了路由，使得特定路由的前缀是

/api/

您必须调用API路由的完整路径：

    $this->call('POST', '/api/article', [
        '_token' => csrf_token(),
        'article_id' => 6,
        'title'=>"hey",
        'body' => "this is a body"
    ])->assertStatus(200);

此外，由于CSRF应该在您的中间件层中实现，并且将

\u token

添加到所有测试请求中既繁琐又愚蠢，因此您可能应该在测试中禁用中间件：

use Illuminate\Foundation\Testing\WithoutMiddleware;

class MyControllerTest {
    use WithoutMiddleware;

    ... public function testmyUnitTest() { ... }
}

---

哦，那真的很有帮助。。。你能帮我纠正我的错误吗？。。。我已将“显示错误”转为“不走运”。@ReyYoung如果您将

$this->call

行替换为我提供的那一行，它将修复您的404错误。