我的 Symfony Web 應(yīng)用程序中有一個(gè) Guard 身份驗(yàn)證。我想執(zhí)行一些單元測(cè)試。我無(wú)法在我的測(cè)試中模擬身份驗(yàn)證。null調(diào)用時(shí)令牌保持不變$tokenStorage->getToken()。筆記：登錄身份驗(yàn)證在dev和環(huán)境下工作prod。我看到很多相關(guān)主題都沒(méi)有成功和文檔。Symfony 版本：3.4。重現(xiàn)：您可以從此repo（symfony 項(xiàng)目）重現(xiàn)錯(cuò)誤。這個(gè) repo 定義了一個(gè)User帶有自定義約束驗(yàn)證器的實(shí)體ExampleValidator。在這個(gè)約束中，我需要擁有當(dāng)前登錄的用戶。代碼示例：手動(dòng)創(chuàng)建用戶后，login測(cè)試中使用的函數(shù)：private function logIn($firewallName = 'main'){ // dummy call to bypass the hasPreviousSession check $crawler = $this->client->request('GET', '/'); $session = $this->client->getContainer()->get('session'); /** @var User $user */ $user = $this->entityManager->getRepository(User::class) ->findOneBy(['email' => 'user1@example.com']); // you may need to use a different token class depending on your application. // for example, when using Guard authentication you must instantiate PostAuthenticationGuardToken $token = new PostAuthenticationGuardToken($user, $firewallName, [new Role('ROLE_CLIENT')]); self::$kernel->getContainer()->get('security.token_storage')->setToken($token); $session->set('_security_'.$firewallName, serialize($token)); $session->save(); $cookie = new Cookie($session->getName(), $session->getId()); $this->client->getCookieJar()->set($cookie);}tokenStorage來(lái)自（來(lái)自服務(wù)功能）的用戶調(diào)用：class ExampleValidator extends ConstraintValidator{ protected $requestStack; protected $em; protected $user_id; public function __construct(RequestStack $request, EntityManager $em, TokenStorage $tokenStorage){ $this->requestStack = $request; $this->em = $em; /** @var User $user */ // Token is always null $user = $tokenStorage->getToken()->getUser(); $this->user_id = $user != "anon." ? $user->getId() : null; } /** * @param $value * @param Constraint $constraint */ public function validate($value, Constraint $constraint) { // validation rules ... }}
1 回答
慕田峪7331174
TA貢獻(xiàn)1828條經(jīng)驗(yàn) 獲得超13個(gè)贊
我相信您的問(wèn)題的根源是您正在使用多個(gè)容器實(shí)例。特別是，您的logIn()函數(shù)在客戶端的容器上運(yùn)行，但驗(yàn)證器來(lái)自您在setUp(). 因此，您logIn()對(duì)客戶端容器所做的更改不會(huì)影響您實(shí)際測(cè)試的驗(yàn)證器。
在任何地方使用相同的容器，例如來(lái)自客戶端的容器，應(yīng)該可以解決這個(gè)問(wèn)題。對(duì)存儲(chǔ)庫(kù)的以下更改使測(cè)試通過(guò)：
diff --git a/tests/AppBundle/Validator/UserTest.php b/tests/AppBundle/Validator/UserTest.php
index f15c854..603e566 100644
--- a/tests/AppBundle/Validator/UserTest.php
+++ b/tests/AppBundle/Validator/UserTest.php
@@ -44,10 +44,7 @@ class UserTest extends WebTestCase{
$this->container = $this->client->getContainer();
$this->entityManager = $this->container->get('doctrine.orm.entity_manager');
- // Set validator
- $kernel = $this->createKernel();
- $kernel->boot();
- $this->validator = $kernel->getContainer()->get('validator');
+ $this->validator = $this->client->getContainer()->get('validator');
// Create one user
$this->createOneUser();
@@ -100,7 +97,7 @@ class UserTest extends WebTestCase{
// you may need to use a different token class depending on your application.
// for example, when using Guard authentication you must instantiate PostAuthenticationGuardToken
$token = new PostAuthenticationGuardToken($user, $firewallName, [new Role('ROLE_CLIENT')]);
- self::$kernel->getContainer()->get('security.token_storage')->setToken($token);
+ $this->client->getContainer()->get('security.token_storage')->setToken($token);
$session->set('_security_'.$firewallName, serialize($token));
$session->save();
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