thinkphp - 请问hyperf如何像Thinkphp框架那样通过HttpResponseException那种捕获异常输出到客户端？

Question

请问hyperf如何像Thinkphp框架那样通过HttpResponseException那种捕获异常输出到客户端？

如：我需要返回给客户端的Json是[code=> 1, data=> ['username'=> '张三'], 'msg'=> 'success']这种格式的，如何在hyperf当中通过像Thinkphp框架那样直接从HttpResponseException捕获异常输出到客户端，而不是还需要 return $this->response->json([code=> 1, data=> ['username'=> '张三'], 'msg'=> 'success'])这种格式输出；

Answer 1

自定义异常类

namespace App\Exception;use Hyperf\Server\Exception\ServerException;class HttpResponseException extends ServerException{    protected $data;    public function __construct($data, $message = "", $code = 0)    {        parent::__construct($message, $code);        $this->data = $data;    }    public function getData()    {        return $this->data;    }}

新建异常处理器：

namespace App\Exception\Handler;use App\Exception\HttpResponseException;use Hyperf\ExceptionHandler\ExceptionHandler;use Hyperf\HttpMessage\Stream\SwooleStream;use Psr\Http\Message\ResponseInterface;use Throwable;class HttpResponseExceptionHandler extends ExceptionHandler{    public function handle(Throwable $throwable, ResponseInterface $response)    {        // 判断异常是不是 HttpResponseException        if ($throwable instanceof HttpResponseException) {            // 获取异常里的数据            $data = $throwable->getData();            // 转成 JSON 响应            $response = $response->withAddedHeader('content-type', 'application/json')->withBody(new SwooleStream(json_encode($data)));        }        return $response;    }    public function isValid(Throwable $throwable): bool    {        return $throwable instanceof HttpResponseException;    }}

在控制器里：

use App\Exception\HttpResponseException;public function someAction(){    //...    // 抛出自定义异常    throw new HttpResponseException(['code' => 1, 'data' => ['username' => '张三'], 'msg' => 'success']);}