WebSocket握手期间出错：意外的响应代码：400 Spring boot websockets

Posted 2023-03-04

【中文标题】WebSocket握手期间出错：意外的响应代码：400 Spring boot websockets

【英文标题】：Error during WebSocket handshake: Unexpected response code: 400 Spring boot websockets

【发布时间】：2021-04-09 18:59:10

【问题描述】：

我正在尝试实现 websocket。我的代码：

在 Spring Security 中，我允许使用身份验证的路径

@Override
protected void configure(HttpSecurity httpSecurity) throws Exception 
    httpSecurity.csrf()
            .disable()
            .authorizeRequests()
            .antMatchers("/register","/login","/testAuth","/web","/ws").permitAll()
            .anyRequest().authenticated()
            .and()
            .exceptionHandling()
            .authenticationEntryPoint(jwtAuthenticationEntryPoint)
            .and()
            .sessionManagement()
            .sessionCreationPolicy(SessionCreationPolicy.STATELESS);
    httpSecurity.addFilterBefore(jwtRequestFilter, UsernamePasswordAuthenticationFilter.class);

以及为 ws 配置

@Configuration
@EnableWebSocket
public class WebSocketConfig implements WebSocketConfigurer 


    @Override
    public void registerWebSocketHandlers(WebSocketHandlerRegistry registry) 
        registry.addHandler(new WsMessageHandler(), "/ws")
                .addInterceptors(new CustomInterceptor())
                .setAllowedOrigins("*")
                .setHandshakeHandler(new DefaultHandshakeHandler())
                .withSockJS();
    

`

  public class CustomInterceptor implements HandshakeInterceptor 
        @Override
        public boolean beforeHandshake(ServerHttpRequest serverHttpRequest, ServerHttpResponse serverHttpResponse, WebSocketHandler webSocketHandler, Map<String, Object> map) throws Exception 
            return true;
        
    
        @Override
        public void afterHandshake(ServerHttpRequest serverHttpRequest, ServerHttpResponse serverHttpResponse, WebSocketHandler webSocketHandler, @Nullable Exception e) 
    
        
    

作为客户端，我使用简单的 js 代码：

var exampleSocket = new WebSocket("ws://localhost:8080/ws", "protocolOne");
           exampleSocket.onopen = function() 
              
              // Web Socket is connected, send data using send()
              ws.send("Message to send");
              alert("Message is sent...");
           ;

但是我收到错误：

到 'ws://localhost:8080/ws' 的 WebSocket 连接失败：期间出错 WebSocket 握手：意外响应代码：400

这是什么原因造成的？一切都应该是正确的。谢谢解答

【参考方案1】：

我发现您的代码有一些错误：

如果你使用 WebSocket() 作为客户端，你应该在 WebSocketConfig 中注释 withSocktJS()。并喜欢它

@Configuration
@EnableWebSocket
public class WebSocketConfig2 implements WebSocketConfigurer 

    @Override
    public void registerWebSocketHandlers(WebSocketHandlerRegistry registry) 
        registry.addHandler(new WsMessageHandler(), "/ws")
                .addInterceptors(new CustomInterceptor())
                .setAllowedOrigins("*")
                .setHandshakeHandler(new DefaultHandshakeHandler())
//                .withSockJS()
        ;
    

您的客户端 js 代码错误。没有ws。您应该使用 exampleSocket 发送。如下所示

                var exampleSocket = new WebSocket("ws://localhost:8081/ws", "protocolOne");
                exampleSocket.onopen = function(ws) 
                    console.log('-----------------', ws);
                    // Web Socket is connected, send data using send()
                    exampleSocket.send("Message to send");
                    alert("Message is sent...");
                ;

在客户端发送 websocket 时，您指向一个名为“protocolOne”的协议，您应该在 CustomInterceptor 中将其返回。您可以执行以下操作。

public class CustomInterceptor implements HandshakeInterceptor 
    @Override
    public boolean beforeHandshake(ServerHttpRequest serverHttpRequest, ServerHttpResponse serverHttpResponse, WebSocketHandler webSocketHandler, Map<String, Object> map) throws Exception 
        return true;
    

    @Override
    public void afterHandshake(ServerHttpRequest serverHttpRequest, ServerHttpResponse serverHttpResponse, WebSocketHandler webSocketHandler, @Nullable Exception e) 
        if(serverHttpRequest.getHeaders() == null )
            return ;
        if(serverHttpRequest.getHeaders().get("Sec-WebSocket-Protocol") == null)
            return ;
        String protocol = (String) serverHttpRequest.getHeaders().get("Sec-WebSocket-Protocol").get(0);
        if(protocol == null)
            return ;

        serverHttpResponse.getHeaders().add("Sec-WebSocket-Protocol", protocol);