我正在使用FormRequest来验证智能手机应用程序在API调用中发送的内容。所以,我希望FormRequest在验证失败时总是返回json。
我看到了Laravel框架的以下源代码,如果reqeust是Ajax或wantJson,则FormRequest的默认行为是返回json。
//Illuminate\Foundation\Http\FormRequest class
/**
* Get the proper failed validation response for the request.
*
* @param array $errors
* @return \Symfony\Component\HttpFoundation\Response
*/
public function response(array $errors)
{
if ($this->ajax() || $this->wantsJson()) {
return new JsonResponse($errors, 422);
}
return $this->redirector->to($this->getRedirectUrl())
->withInput($this->except($this->dontFlash))
->withErrors($errors, $this->errorBag);
}
我知道我可以在请求标头中添加Accept= application/json
。 FormRequest将返回json。但我希望提供一种更简单的方法来通过默认支持json来请求我的API,而无需设置任何标头。所以,我试图在Illuminate\Foundation\Http\FormRequest
类中找到一些强制FormRequest响应json的选项。但我没有找到任何默认支持的选项。
我试图覆盖我的应用程序请求抽象类,如下所示:
<?php
namespace Laravel5Cg\Http\Requests;
use Illuminate\Foundation\Http\FormRequest;
use Illuminate\Http\JsonResponse;
abstract class Request extends FormRequest
{
/**
* Force response json type when validation fails
* @var bool
*/
protected $forceJsonResponse = false;
/**
* Get the proper failed validation response for the request.
*
* @param array $errors
* @return \Symfony\Component\HttpFoundation\Response
*/
public function response(array $errors)
{
if ($this->forceJsonResponse || $this->ajax() || $this->wantsJson()) {
return new JsonResponse($errors, 422);
}
return $this->redirector->to($this->getRedirectUrl())
->withInput($this->except($this->dontFlash))
->withErrors($errors, $this->errorBag);
}
}
如果我们需要强制响应json,我将protected $forceJsonResponse = false;
添加到设置中。并且,在每个FormRequest中,它从Request抽象类扩展而来。我设定了这个选项。
例如:我创建了一个StoreBlogPostRequest并为此FormRequest设置$forceJsoResponse=true
并使其响应json。
<?php
namespace Laravel5Cg\Http\Requests;
use Laravel5Cg\Http\Requests\Request;
class StoreBlogPostRequest extends Request
{
/**
* Force response json type when validation fails
* @var bool
*/
protected $forceJsonResponse = true;
/**
* Determine if the user is authorized to make this request.
*
* @return bool
*/
public function authorize()
{
return true;
}
/**
* Get the validation rules that apply to the request.
*
* @return array
*/
public function rules()
{
return [
'title' => 'required|unique:posts|max:255',
'body' => 'required',
];
}
}
我构建了一个类似以下的中间件:
namespace Laravel5Cg\Http\Middleware;
use Closure;
use Symfony\Component\HttpFoundation\HeaderBag;
class AddJsonAcceptHeader
{
/**
* Add Json HTTP_ACCEPT header for an incoming request.
*
* @param \Illuminate\Http\Request $request
* @param \Closure $next
* @return mixed
*/
public function handle($request, Closure $next)
{
$request->server->set('HTTP_ACCEPT', 'application/json');
$request->headers = new HeaderBag($request->server->getHeaders());
return $next($request);
}
}
这是工作。但我想知道这个解决方案好吗?在这种情况下,有没有Laravel Way可以帮助我?
答案 0 :(得分:29)
令我难以理解为什么在拉拉维尔这么难做。最后,根据你的想法覆盖Request类,我想出了这个。
app/Http/Requests/ApiRequest.php
<?php
namespace App\Http\Requests;
class ApiRequest extends Request
{
public function wantsJson()
{
return true;
}
}
然后,在每个控制器中只传递\App\Http\Requests\ApiRequest
public function index(ApiRequest $request)
答案 1 :(得分:27)
我知道这篇文章有点陈旧,但我只是制作了一个用“application / json”替换请求的“Accept”标题的中间件。这使得wantsJson()
函数在使用时返回true
。 (这在Laravel 5.2中进行了测试,但我认为它在5.1中的工作原理相同)
以下是您实施该方法的方法:
创建文件app/Http/Middleware/Jsonify.php
namespace App\Http\Middleware;
use Closure;
class Jsonify
{
/**
* Change the Request headers to accept "application/json" first
* in order to make the wantsJson() function return true
*
* @param \Illuminate\Http\Request $request
* @param \Closure $next
*
* @return mixed
*/
public function handle($request, Closure $next)
{
$request->headers->set('Accept', 'application/json');
return $next($request);
}
}
将中间件添加到$routeMiddleware
文件的app/Http/Kernel.php
表中
protected $routeMiddleware = [
'auth' => \App\Http\Middleware\Authenticate::class,
'guest' => \App\Http\Middleware\RedirectIfAuthenticated::class,
'jsonify' => \App\Http\Middleware\Jsonify::class
];
最后在您的routes.php
中使用它,就像使用任何中间件一样。就我而言,它看起来像这样:
Route::group(['prefix' => 'api/v1', 'middleware' => ['jsonify']], function() {
// Routes
});
答案 2 :(得分:2)
如果您的请求具有 X-Request-With:XMLHttpRequest 标头或接受内容类型为application / json ,则FormRequest将自动返回包含错误的json响应,并带有状态为422。
答案 3 :(得分:0)
我只是覆盖failedValidation
函数
protected function failedValidation(Validator $validator)
{
if ($this->wantsJson()) {
// flatten all the message
$collection = collect($validator->errors())->flatten()->values()->all();
throw new HttpResponseException(Response::error('Validation Error', $collection));
}
parent::failedValidation($validator);
}
因此输出示例:
{
"error": true,
"message": "Validation Error",
"reference": [
"The device id field is required.",
"The os version field is required.",
"The apps version field is required."
],
}
答案 4 :(得分:0)
基于ZeroOne's response,如果您使用的是Form Request validation,则可以覆盖failedValidation方法以在验证失败的情况下始终返回json。
此解决方案的优点是,您不会覆盖所有返回json的响应,而只是覆盖验证失败。因此,对于所有其他Php异常,您仍然会看到友好的Laravel错误页面。
namespace App\Http\Requests;
use Illuminate\Contracts\Validation\Validator;
use Illuminate\Foundation\Http\FormRequest;
use Illuminate\Http\Exceptions\HttpResponseException;
use Symfony\Component\HttpFoundation\Response;
class InventoryRequest extends FormRequest
{
protected function failedValidation(Validator $validator)
{
throw new HttpResponseException(response($validator->errors(), Response::HTTP_UNPROCESSABLE_ENTITY));
}