FOSRestBundle，生成創建資源的URL

Question

我正在使用Symfony和FOSRestBundle創建一個REST API，而且這兩者都非常新穎。

現在我想知道如何生成我剛剛創建的資源的URL。路線都設置這樣的：

# 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 

人物控制器，創建資源的一部分：

$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：全控制器代碼：

<?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 { 

     } 
    } 
} 

的不像我預期的那樣工作的是generateUrl部分設置位置標題。它吐出http://optc.local/api/characters？id = 2。這當然只會列出全部資源。但我想要的是http://optc.local/api/characters/2。

我該怎麼做？似乎我錯過了一些簡單的東西。

（順便說一句，關於返回位置頭是從http://williamdurand.fr/2012/08/02/rest-apis-with-symfony2-the-right-way/ PHP的一部分，所以我希望這是「正確」的方式。）

Answer 1

必須使用的，而不是get_characters路線get_character路線，

我建議你是否實現ClassResourceInterface或使用RouteResource註釋，可以使用方法名作爲getAction，cgetAction（這僅僅是一個建議）

Answer 2

你應該在終端檢查app/console debug:router看什麼名字symfony是命名在我的情況下，它使用的減，而不是下劃線

即get-character