Laravel phpunit测试获取参数

Question

我正在为我的控制器编写一些测试,但我的一个测试不起作用.它倾向于搜索并将结果返回到页面.但它实际上是重定向到主页.这是我的代码:

use DatabaseMigrations;
protected $user;
public function setUp()
{
    parent::setUp();

    $this->seed();

    $this->user = factory(User::class)->create(['role_id' => 3]);
}

/** @test */
public function test_manage_search_user()
{
    $response = $this->followingRedirects()->actingAs($this->user)->get('/manage/users/search', [
        'choices' => 'username',
        'search' => $this->user->username,
    ]);

    $response->assertViewIs('manage.users');
    $response->assertSuccessful();
    $response->assertSee($this->user->email);
}

您应该使其工作的URL如下所示:

http://localhost/manage/users/search?choices=username&search=Test

我再次检查,看起来它没有在带有get请求的参数中给出.我怎样才能解决这个问题？

Answer 1

我有同样的问题试图测试GET请求,你实际上不能传递参数$this->get('uri', [header])但你可以通过使用$this->call,如果你签入MakesHttpRequests.php你可以看到,this->get()实际上是使用调用方法.

通过向get方法添加数组,您将更改请求标头,这就是您没有获取参数的原因.

public function get($uri, array $headers = [])
{
    $server = $this->transformHeadersToServerVars($headers);

    return $this->call('GET', $uri, [], [], [], $server);
}

public function call($method, $uri, $parameters = [], $cookies = [], $files = [], $server = [], $content = null)
{
    $kernel = $this->app->make(HttpKernel::class);

    $files = array_merge($files, $this->extractFilesFromDataArray($parameters));

    $symfonyRequest = SymfonyRequest::create(
        $this->prepareUrlForRequest($uri), $method, $parameters,
        $cookies, $files, array_replace($this->serverVariables, $server), $content
    );

    $response = $kernel->handle(
        $request = Request::createFromBase($symfonyRequest)
    );

    if ($this->followRedirects) {
        $response = $this->followRedirects($response);
    }

    $kernel->terminate($request, $response);

    return $this->createTestResponse($response);
}

因此,如果您要测试GET请求,则必须执行以下操作:

$request = $this->call('GET', '/myController', ["test"=>"test"]);

在您的控制器中,您应该能够获得这样的参数:

public function myController(Request $request)
{
    $requestContent = $request->all();
    $parameter = $requestContent['test'];
}

Answer 2

我正在使用Laravel 5.X（更确切地说是5.6），您可以使用以下方式传递自定义参数：

 $response = $this->json('GET',  '/url/endpoint',['params'=>'value']);

Answer 3

您可以使用路由助手构建带有查询字符串的 url。在你的情况下，我会做这样的事情。假设路由名称是 manage.users.search

$route = route('manage.users.search', [
    'choices'=> 'username',
    'search' => $this->user->username,
]);

$response = $this->followingRedirects()
    ->actingAs($this->user)
    ->get($route);