api - 在yii2 REST API中自定义错误消息

Question

我想自定义以下错误消息

Error 404: Not Found
{
    "name": "Not Found",
    "message": "Object not found: 6",
    "code": 0,
    "status": 404,
    "type": "yii\\web\\NotFoundHttpException"
}

至:

Error 404: Not Found
{
    "name": "Not Found",
    "message": "Country not found for: 6",
    "code": 404,
    "status": Error
}

需要在哪里编写此定制代码？

Answer 1

您需要覆盖扩展yii \ web \ ErrorHandler的errorHandler，只是convertExceptionToArray方法

/**
 * Converts an exception into an array.
 * @param \Exception|\Error $exception the exception being converted
 * @return array the array representation of the exception.
 */
protected function convertExceptionToArray($exception)
{
    if (!YII_DEBUG && !$exception instanceof UserException && !$exception instanceof HttpException) {
        $exception = new HttpException(500, Yii::t('yii', 'An internal server error occurred.'));
    }

    $array = [
        'name' => ($exception instanceof Exception || $exception instanceof ErrorException) ? $exception->getName() : 'Exception',
        'message' => $exception->getMessage(),
        'code' => $exception->getCode(),
    ];
    if ($exception instanceof HttpException) {
        $array['status'] = $exception->statusCode;
    }
    if (YII_DEBUG) {
        $array['type'] = get_class($exception);
        if (!$exception instanceof UserException) {
            $array['file'] = $exception->getFile();
            $array['line'] = $exception->getLine();
            $array['stack-trace'] = explode("\n", $exception->getTraceAsString());
            if ($exception instanceof \yii\db\Exception) {
                $array['error-info'] = $exception->errorInfo;
            }
        }
    }
    if (($prev = $exception->getPrevious()) !== null) {
        $array['previous'] = $this->convertExceptionToArray($prev);
    }

    return $array;
}

上面的代码来自Yii，您只需要对其稍作调整。
然后将其添加到配置中:(通常是web.php)

    'errorHandler' => [
        'class' => 'your\namespace\YourErrHandler',
    ],