[英]FOSRestBundle: How to set an array to an URL in my Rest Api
[英]Set up registration FOSUserBundle with FOSRestBundle REST API
問題已解決,請檢查我的答案。
我正在我的Symfony2.7 rest api上建立一個注冊端點。 我正在使用FosRestBundle和FosUserBundle
這是用戶模型:
<?php
namespace AppBundle\Entity;
use FOS\UserBundle\Model\User as BaseUser;
use Doctrine\ORM\Mapping as ORM;
/**
* @ORM\Entity
* @ORM\Table(name="fos_user")
*/
class User extends BaseUser {
/**
* @ORM\Id
* @ORM\Column(type="integer")
* @ORM\GeneratedValue(strategy="AUTO")
*/
protected $id;
public function __construct() {
parent::__construct();
// your own logic
}
}
\\ 這是UserType表單: \\
class UserType extends AbstractType
{
/**
* @param FormBuilderInterface $builder
* @param array $options
*/
public function buildForm(FormBuilderInterface $builder, array $options)
{
$builder
->add('email', 'email')
->add('username', null)
->add('plainPassword', 'repeated', array(
'type' => 'password',
'first_options' => array('label' => 'password'),
'second_options' => array('label' => 'password_confirmation'),
))
;
}
/**
* @param OptionsResolverInterface $resolver
*/
public function setDefaultOptions(OptionsResolverInterface $resolver)
{
$resolver->setDefaults(array(
'data_class' => 'AppBundle\Entity\User',
'csrf_protection' => false,
));
}
/**
* @return string
*/
public function getName()
{
return 'user';
}
}
而這個帖子用戶控制器:
public function postUserAction(\Symfony\Component\HttpFoundation\Request $request) {
$user = new \AppBundle\Entity\User();
$form = $this->createForm(new \AppBundle\Form\UserType(), $user);
$form->handleRequest($request);
if ($form->isValid()) {
$em = $this->getDoctrine()->getManager();
$em->persist($user);
$em->flush();
$view = $this->view(array('token'=>$this->get("lexik_jwt_authentication.jwt_manager")->create($user)), Codes::HTTP_CREATED);
return $this->handleView($view);
}
return array(
'form' => $form,
);
}
問題是,當我提交錯誤的信息或空信息時,服務器返回一個錯誤的格式化500錯誤,其中帶有錯誤格式化條目列表的json響應狀態中非空行的空值的doctrine / mysql詳細信息。
有關如何解決此問題的任何想法? 為什么驗證得到通過和
好好花了很多時間閱讀FOSUserBundle代碼,特別是注冊控制器和表格工廠,我想出了一個完全可行的解決方案。
在做任何事情之前,不要忘記在symfony2配置中禁用CSRF。
這是我用來注冊的控制器:
public function postUserAction(\Symfony\Component\HttpFoundation\Request $request) {
/** @var $formFactory \FOS\UserBundle\Form\Factory\FactoryInterface */
$formFactory = $this->get('fos_user.registration.form.factory');
/** @var $userManager \FOS\UserBundle\Model\UserManagerInterface */
$userManager = $this->get('fos_user.user_manager');
/** @var $dispatcher \Symfony\Component\EventDispatcher\EventDispatcherInterface */
$dispatcher = $this->get('event_dispatcher');
$user = $userManager->createUser();
$user->setEnabled(true);
$event = new \FOS\UserBundle\Event\GetResponseUserEvent($user, $request);
$dispatcher->dispatch(\FOS\UserBundle\FOSUserEvents::REGISTRATION_INITIALIZE, $event);
if (null !== $event->getResponse()) {
return $event->getResponse();
}
$form = $formFactory->createForm();
$form->setData($user);
$form->handleRequest($request);
if ($form->isValid()) {
$event = new \FOS\UserBundle\Event\FormEvent($form, $request);
$dispatcher->dispatch(\FOS\UserBundle\FOSUserEvents::REGISTRATION_SUCCESS, $event);
$userManager->updateUser($user);
if (null === $response = $event->getResponse()) {
$url = $this->generateUrl('fos_user_registration_confirmed');
$response = new \Symfony\Component\HttpFoundation\RedirectResponse($url);
}
$dispatcher->dispatch(\FOS\UserBundle\FOSUserEvents::REGISTRATION_COMPLETED, new \FOS\UserBundle\Event\FilterUserResponseEvent($user, $request, $response));
$view = $this->view(array('token' => $this->get("lexik_jwt_authentication.jwt_manager")->create($user)), Codes::HTTP_CREATED);
return $this->handleView($view);
}
$view = $this->view($form, Codes::HTTP_BAD_REQUEST);
return $this->handleView($view);
}
現在棘手的部分是使用REST提交表單。 問題是,當我發送像這樣的JSON時:
{
"email":"xxxxx@xxxx.com",
"username":"xxx",
"plainPassword":{
"first":"xxx",
"second":"xxx"
}
}
API沒有提交任何內容就響應了。
解決方案是Symfony2正在等待您將表單數據封裝在表單名稱中!
問題是“我沒有創建這種形式所以我不知道它的名字是什么......”。 所以我再次使用捆綁代碼,發現表單類型為fos_user_registration,getName函數返回fos_user_registration_form。
結果我嘗試以這種方式提交我的JSON:
{"fos_user_registration_form":{
"email":"xxxxxx@xxxxxxx.com",
"username":"xxxxxx",
"plainPassword":{
"first":"xxxxx",
"second":"xxxxx"
}
}}
瞧! 有效。 如果你正在努力設置你的fosuserbundle與fosrestbundle和LexikJWTAuthenticationBundle只是問我,我會很樂意提供幫助。
另一種方式是沒有來自FOSUserBundle的表單進行注冊。 使用params發出POST請求:電子郵件,用戶,密碼。
public function postUserAction(Request $request)
{
$userManager = $this->get('fos_user.user_manager');
$email = $request->request->get('email');
$username = $request->request->get('user');
$password = $request->request->get('password');
$email_exist = $userManager->findUserByEmail($email);
$username_exist = $userManager->findUserByUsername($username);
if($email_exist || $username_exist){
$response = new JsonResponse();
$response->setData("Username/Email ".$username."/".$email." existiert bereits");
return $response;
}
$user = $userManager->createUser();
$user->setUsername($username);
$user->setEmail($email);
$user->setLocked(false);
$user->setEnabled(true);
$user->setPlainPassword($password);
$userManager->updateUser($user, true);
$response = new JsonResponse();
$response->setData("User: ".$user->getUsername()." wurde erstellt");
return $response;
}
@Adel'Sean'Helal你的方式不起作用,至少使用FOSRestBundle,FOSUserBundle和Symfony的最新版本。 我差點把自己射向腦袋試圖讓它發揮作用。 最后我找到了解決方案,而且非常簡單。 只需要解析請求。
我的控制器代碼的片段
...
$form->setData($user);
// THIS LINE DO THE MAGIC
$data = json_decode($request->getContent(), true);
if ($data === null) {
throw new BadRequestHttpException();
}
$form->submit($data);
if ( ! $form->isValid()) {
$event = new FormEvent($form, $request);
$dispatcher->dispatch(FOSUserEvents::REGISTRATION_FAILURE, $event);
if (null !== $response = $event->getResponse()) {
return $response;
}
return new JsonResponse($this->getFormErrors($form), Response::HTTP_BAD_REQUEST);
}
...
composer.json依賴項:
...
"symfony/lts": "^3",
"symfony/flex": "^1.0",
"friendsofsymfony/rest-bundle": "^2.3",
"friendsofsymfony/user-bundle": "^2.0",
"lexik/jwt-authentication-bundle": "^2.4",
...
我的功能測試代碼:
namespace App\Tests\Controller;
use Symfony\Bundle\FrameworkBundle\Test\WebTestCase;
use Symfony\Component\DependencyInjection\Container;
class ApiUserControllerTest extends WebTestCase
{
/**
* @var Container
*/
private $container;
public function setUp()
{
self::bootKernel();
$this->container = self::$kernel->getContainer();
}
public function testRegistration()
{
$userData = [
'username' => 'test',
'email' => 'test@email.com',
'plainPassword' => [
'first' => 'test123', 'second' => 'test123'
]
];
$client = $this->container->get('eight_points_guzzle.client.rest');
$response = $client->post(
'api/registration',
['json' => $userData]
);
$bodyResponse = \GuzzleHttp\json_decode($response->getBody(), true);
$this->assertEquals(201, $response->getStatusCode());
$this->assertArrayHasKey('token', $bodyResponse);
$this->assertNotEmpty($bodyResponse['token']);
}
}
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