laravel-异常处理
laravel 异常抛出
在开发中错误处理是非常重要的，最重要是两点：第一，声明异常的错误代码和信息。第二，不同场景下抛出异常的有所不同，比如 api 中一般是前端 ajax 请求，那么抛出的异常应该是 json 形式，如果是开发模式下，页面中抛出普通异常，便于程序猿调错，在生产环境下，则统一将错误汇总放入一个错误页面进行渲染。
声明异常
就以一个用户模块来举例：
在 /config/exceptions.php 中声明抛出异常的信息和错误代码
<?php return [ 'not_found_api' => ['message' => '接口未找到', 'code' => '404'], 'not_found_page' => ['message' => '页面未找到', 'code' => '404'], 'user' => [ 'already_logout' => ['message' => '您已处于登出状态，请重新登录', 'code' => '40005'], 'permission_deny' => ['message' => '您无访问权限', 'code' => '40001'], 'captcha_error' => ['message' => '短信验证码错误', 'code' => '40002'], 'not_found' => ['message' => '该用户不存在', 'code' => '40003'], 'not_admin' => ['message' => '该账号没有管理后台权限！', 'code' => '40004'], ] ];
在 /app/Exceptions/ 下的文件目录结构：
app/Exceptions
BaseException.php
User.php
Handler.php
BaseException.php : 接受和处理异常，并传入父类的构造函数异常处理中
<?phpnamespace App\Exceptions;use JsonSerializable;use Exception;class BaseException extends Exception implements JsonSerializable{const EXCEPTION_CONFIG_PREFIX = 'exceptions.';public function __construct($config = '', $message = null, $code = null){$exception = config(static::EXCEPTION_CONFIG_PREFIX.$config);$code = is_null($code)? $exception['code'] : $code;$message = is_null($message) ? $exception['message'] : $message;parent::__construct($message, $code);}public function jsonSerialize(){return ['errCode' => $this->getCode(),'message' => $this->getMessage()];}public function setMessage($message){$this->message = $message;}public function setCode($code){$this->code = $code;}}
User.php : 声明用户 User 模块中的异常
<?phpnamespace App\Exceptions;class User extends BaseException{const EXCEPTION_CONFIG_PREFIX = 'exceptions.user.';}
Handler.php : 处理异常
<?phpnamespace App\Exceptions;use Exception;use Illuminate\Database\Eloquent\ModelNotFoundException;use Symfony\Component\HttpKernel\Exception\HttpException;//区别是否 ajax 请求 api 而分别处理的 404 错误use App\Exceptions\NotFoundApi as NotFoundApiException;use App\Exceptions\NotFoundPage as NotFoundPageException;//在此命名空间 /vendor/symfony/http-kernel/Exception 下有许多异常，可以参考一下use Symfony\Component\HttpKernel\Exception\NotFoundHttpException;use Illuminate\Foundation\Exceptions\Handler as ExceptionHandler;use App\Traits\Response as ResponseTrait;class Handler extends ExceptionHandler{use ResponseTrait;/** * A list of the exception types that should not be reported. * * @var array */protected $dontReport = [HttpException::class,ModelNotFoundException::class,];/** * Report or log an exception. * * This is a great spot to send exceptions to Sentry, Bugsnag, etc. * * @param \Exception $e * @return void */public function report(Exception $e){return parent::report($e);}/** * Render an exception into an HTTP response. * * @param \Illuminate\Http\Request $request * @param \Exception $e * @return \Illuminate\Http\Response */public function render($request, Exception $e){//如果是生产环境if (env('APP_DEGUB')) {$exceptions = [];$exceptions['code'] = $e->getCode();$exceptions['message'] = $e->getMessage();//上线时，需要对抛出的异常进行汇总交互至前端 view 中return response()->view("errors.exceptions", ['exceptions' => $exceptions]);}if ($e instanceof ModelNotFoundException) {$e = new NotFoundHttpException($e->getMessage(), $e);}/** * 调试环境下，不需要返回json格式的结果 */if(!env('APP_NEED_JSON_RESPONSE')){return parent::render($request, $e);}if($request->ajax() || $request->wantsJson()){if($e instanceof NotFoundHttpException){$e = new NotFoundApiException;}return $this->jsonResponse(null, $e->getMessage(), $e->getCode());}else{if($e instanceof NotFoundHttpException){$e = new NotFoundPageException;}return parent::render($request, $e);}}}
准备工作都做好后，抛异常就非常简单了 ：
use App\Exceptions\User as UserException;throw new UserException('captcha_error');
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