1

我对 websocket 有问题，特别是 SimpMessagingTemplate 和提供 spring 的 simple-broker（版本 4.3.4.RELEASE）。我在课堂上向用户发送消息：

@Autowired
SimpMessagingTemplate simpMessagingTemplate;

public void onMessage(final TextMessage message, Session session) throws JMSException {
    ....
    this.simpMessagingTemplate.convertAndSendToUser(
                        userInfo.getIp(), "/queue/resp", message.getText());
}

这是我的 websocket 配置：

    <websocket:message-broker application-destination-prefix="/app" user-destination-prefix="/user">
        <websocket:stomp-endpoint path="/workplace">
            <websocket:handshake-interceptors>
                <bean class="org.springframework.web.socket.server.support.HttpSessionHandshakeInterceptor"/>
            </websocket:handshake-interceptors>
            <websocket:sockjs/>
        </websocket:stomp-endpoint>
        <websocket:simple-broker prefix="/topic, /queue, /user"/>
    </websocket:message-broker>

这是我的 web.xml 内容：

<servlet>
    <servlet-name>mvc-dispatcher</servlet-name>
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
    <load-on-startup>1</load-on-startup>
    <async-supported>true</async-supported>
</servlet>

<servlet-mapping>
    <servlet-name>mvc-dispatcher</servlet-name>
    <url-pattern>/rest/*</url-pattern>
</servlet-mapping>

<context-param>
    <param-name>contextConfigLocation</param-name>
    <param-value>/WEB-INF/mvc-dispatcher-servlet.xml</param-value>
</context-param>

<listener>
    <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>

我的控制器：

@Controller 公共类 UserController 扩展 TextWebSocketHandler {

@MessageMapping("/req")
@SendToUser(destinations = "/queue/resp")
public String greeting(String greeting)
        throws Exception {
    System.out.println("greeeeeeeeeeeeeeeeeeeeeeeeeeeeeting: " + greeting);
    return "{ \"date\": "+ new Date().getTime() + ", \"message\": " + greeting + "}";
}

@MessageExceptionHandler
@SendToUser("/queue/errors")
public String handleException(Exception ex) {
    ex.printStackTrace();
    System.out.println(ex);
    return ex.getMessage();
}

}

还有我在客户端的 js 代码：

    <script>
var service = {},  socket = {
client: null,
stomp: null
};

service.userIpAddress = 'someIpAddress';
service.RECONNECT_TIMEOUT = 25000;
service.SOCKET_URL = 'http://localhost:8080/examinee_workplace/rest/workplace/';


service.send = function(message) {
console.info('try send message = ', message);
var id = Math.floor(Math.random() * 1000000);
socket.stomp.send("/app/req", {
priority: 9
}, JSON.stringify({
message: message,
id: id
}));
};

var startListener = function() {
console.info('subscribe');
socket.stomp.subscribe('/user/' + service.userIpAddress + '/queue/resp', function(resp) {
console.warn('resp = ', resp);
});
socket.stomp.subscribe('/user/' + service.userIpAddress + '/queue/errors', function(resp) {
var body = JSON.parse(resp.body);
console.warn('error body = ', body);
});
};

var reconnect = function() {
console.info('reconnect. this.RECONNECT_TIMEOUT = ', this.RECONNECT_TIMEOUT);
setTimeout(service.initialize,1000);
};

service.initialize = function() {
socket.client = new SockJS(service.SOCKET_URL);
socket.stomp = Stomp.over(socket.client);
socket.stomp.connect({}, startListener);
socket.stomp.onclose = reconnect;
};

service.initialize();

最后是我的问题 - 我在客户端上收到消息，但不是全部。某些消息未传递到客户端（防病毒卡巴斯基已关闭）。你有什么想法为什么？

4

1 回答 1

0

我只是用 JavaConfig 替换 Spring XML Configuration。我在这里找到了答案：Stomp over socket using sockjs can't connect with Spring 4 WebSocket

于 2017-03-15T14:58:35.360 回答