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关于Hyperf框架websocket连接建立报错的疑问

### 问题描述
因想体验下hyperf的websocket功能，故照着官方文档（https://hyperf.wiki/#/zh-cn/websocket-server）
copy了一下代码，结果在浏览器客户端建立连接时后台日志报错，显示如下：

PHP Fatal error:  Uncaught TypeError: Argument 2 passed to Hyperf\WebSocketServer\Exception\Handler\WebSocketExceptionHandler::handle() must implement interface Psr\Http\Message\ResponseInterface, null given, called in /home/web/hyperf-skeleton/vendor/hyperf/exception-handler/src/ExceptionHandlerDispatcher.php on line 60 and defined in /home/web/hyperf-skeleton/vendor/hyperf/websocket-server/src/Exception/Handler/WebSocketExceptionHandler.php:39

大致意思是Hyperf\WebSocketServer\Exception\Handler\WebSocketExceptionHandler  类下的handle方法第二个参数传入了null
即
Hyperf\WebSocketServer\Exception\Handler\WebSocketExceptionHandler->handle(Object(Error), NULL)
因这个错误报的我毫无头绪，故想询问一下

### Swoole版本，PHP版本，以及操作系统版本信息
PHP 7.3.12
swoole 4.5.0-alpha

### 实际看到的错误信息又是什么？
[INFO] TaskWorker#5 started.
[INFO] TaskWorker#6 started.
[INFO] TaskWorker#7 started.
[INFO] TaskWorker#8 started.
[INFO] Worker#0 started.
[INFO] WebSocket Server listening at 0.0.0.0:9502
[INFO] HTTP Server listening at 0.0.0.0:9501
[INFO] TaskWorker#1 started.
[INFO] TaskWorker#3 started.
[INFO] TaskWorker#4 started.
[INFO] TaskWorker#2 started.
PHP Fatal error:  Uncaught TypeError: Argument 2 passed to Hyperf\WebSocketServer\Exception\Handler\WebSocketExceptionHandler::handle() must implement interface Psr\Http\Message\ResponseInterface, null given, called in /home/web/hyperf-skeleton/vendor/hyperf/exception-handler/src/ExceptionHandlerDispatcher.php on line 60 and defined in /home/web/hyperf-skeleton/vendor/hyperf/websocket-server/src/Exception/Handler/WebSocketExceptionHandler.php:39

发布于4年前 · 27 次浏览 · 来自提问