我正在使用Symfony和FOSRestBundle创建REST API,而且我对这两者都很陌生。
现在我想知道如何生成我刚刚创建的资源的URL。路线设置如下:
# app/config/routing.yml
characters:
type: rest
prefix: /api
resource: "@Optc/Resources/config/routing/characters_routing.yml"
NelmioApiDocBundle:
prefix: /api/doc
resource: "@NelmioApiDocBundle/Resources/config/routing.yml"
# Resources/Optc/Resources/config/routing/characters_routing.yml
characters:
type: rest
resource: Optc\Controller\CharactersController
创建资源的角色控制器部分:
$character = new Character();
$form = $this->createForm(new CharacterType(), $character);
$form->bind($data);
if ($form->isValid()) {
$em = $this->getDoctrine()->getManager();
$em->persist($character);
$em->flush();
$response->headers->set('Location', $this->generateUrl('get_characters', array('id' => $user->getId()), true));
$view = $this->view($character, 200);
return $this->handleView($view);
}
更新:完整控制器代码:
<?php
namespace Optc\Controller;
use FOS\RestBundle\Controller\Annotations\QueryParam;
use FOS\RestBundle\Controller\FOSRestController;
use Nelmio\ApiDocBundle\Annotation\ApiDoc;
use Optc\Entity\Character;
use Optc\Form\CharacterType;
use Optc\HttpFoundation\File\Base64File;
use Symfony\Component\HttpFoundation\Request;
use Symfony\Component\HttpKernel\Exception\HttpException;
/**
* Characters Controller
*/
class CharactersController extends FOSRestController
{
/**
* Get the list of characters.
*
* @param string $page integer with the page number (requires param_fetcher_listener: force)
*
* @return array data
*
* @QueryParam(name="page", requirements="\d+", default="1", description="Page number of the overview.")
* @ApiDoc()
*/
public function getCharactersAction($page)
{
$characters = $this
->getDoctrine()
->getRepository('Optc:Character')
->findAll();
$view = $this->view($characters, 200);
return $this->handleView($view);
}
public function getCharacterAction($id)
{
$character = $this
->getDoctrine()
->getRepository('Optc:Character')
->findOneById($id);
if (!$character) {
throw new HttpException(404, sprintf('Character with id %d not found!', $id));
}
$view = $this->view($character, 200);
return $this->handleView($view);
}
/**
* Create a new character.
*
* @param Request $request
*
* @return View view instance
*
* @ApiDoc()
*/
public function postCharacterAction(Request $request)
{
$data = $request->request->all();
// If the request contains image date, first convert it from its base64 enconding to a real file
if ($request->request->has('image') && $request->request->get('id')) {
$imagePath = realpath($this->get('kernel')->getRootDir() . '/../web'.$this->container->getParameter('upload_path_characters')).'/'.$request->request->get('id');
$file = Base64File::create($imagePath, $request->request->get('image'));
$data['image'] = $file;
}
$character = new Character();
$form = $this->createForm(new CharacterType(), $character);
$form->bind($data);
var_dump($form->isValid());
var_dump($form->getErrorsAsString());
var_dump($this->generateUrl('get_characters', array('id' => $character->getId()), true));
die();
if ($form->isValid()) {
$em = $this->getDoctrine()->getManager();
$em->persist($character);
$em->flush();
$response->headers->set('Location', $this->generateUrl('acme_demo_user_get', array('id' => $user->getId()), true));
$view = $this->view($character, 200);
return $this->handleView($view);
}
else {
}
}
}
不能像我预期的那样工作的是设置 Location 标头的generateUrl
部分。它吐出http://optc.local/api/characters ?id = 2 。这当然只会列出所有资源。但我想要的是http://optc.local/api/characters / 2 。
我该怎么做?好像我错过了一些简单的东西。
(顺便说一句,关于返回 Location 标题的PHP部分来自http://williamdurand.fr/2012/08/02/rest-apis-with-symfony2-the-right-way/,所以我希望这是&#34;正确的方式。)< / p>
答案 0 :(得分:1)
您必须使用get_character
路线代替get_characters
路线
我建议您实施ClassResourceInterface
或使用RouteResource
注释,可以使用方法名称getAction
,cgetAction
(这只是建议)
答案 1 :(得分:1)
你应该检查终端中的app/console debug:router
以查看symfony命名路由的名称
在我的情况下,它使用了减号而不是下划线
即get-character