解决hyperf中方法需要实现接口的问题

孟帅
2021-02-24 17:19:00
PHP
PHP swoole hyperf

报错内容：

PHP Fatal error: Uncaught TypeError: Argument 1 passed to Hyperf\HttpServer\ResponseEmitter::emit() must implement interface Psr\Http\Message\ResponseInterface, instance of Hyperf\ViewEngine\View given, called in /www/wwwroot/server/vendor/hyperf/http-server/src/Server.php on line 128 and defined in /www/wwwroot/server/vendor/hyperf/http-server/src/ResponseEmitter.php:24

报错代码：

    use Psr\Http\Message\ResponseInterface;
    use function Hyperf\ViewEngine\view;
    
    public function handle(ResponseInterface $response)
    {
        return view('/Common/Response/error', [
             'code' => 500,
             'msg'  => "发生错误"
             'data' => [],
        ]);
    }

分析：在hyperf开发过程中会遇到方法或类需要实现接口的问题，这说明报错的方法或类强制要求对象指针实现相关接口

解决方法有两种情况：

1.已经引入需实现的接口，此时只需用其返回即可

    use Psr\Http\Message\ResponseInterface;
    use Hyperf\HttpMessage\Stream\SwooleStream;
    use function Hyperf\ViewEngine\view;
        
    public function handle(ResponseInterface $response)
    {
         return $response
           ->withHeader("Content-Type", "text/html;charset=utf-8")
           ->withStatus(200)
           ->withBody(new SwooleStream((string) view('/Common/Response/error', [
              'code' => 500,
              'msg'  => "发生错误"
              'data' => [],
       ])));
    }

2.未引入，利用协程切换到需实现的接口，进行相关业务调用即可

    use Psr\Http\Message\ResponseInterface;
    use Hyperf\HttpMessage\Stream\SwooleStream;
    use Hyperf\Utils\Context;
    use function Hyperf\ViewEngine\view;
    
    public function handle()
    {
        return Context::get(ResponseInterface::class)
           ->withHeader("Content-Type", "text/html;charset=utf-8")
           ->withStatus(200)
           ->withBody(new SwooleStream((string) view('/Common/Response/error', [
            'code' => 500,
            'msg'  => "发生错误"
            'data' => [],
        ])));
    }