1
我在Spring Boot应用程序中使用@SubscribeMapping,该应用程序严重依赖于WebSockets进行数据交换。该应用程序使用Spring Security进行保护。使用Spring @SubscribeMapping获取当前用户
客户端我用践踏过的WebSocket:
this.socket = new WebSocket(this.socketUrl);
this.stompClient = Stomp.over(this.socket);
this.stompClient.debug = null;
this.stompClient.connect({},
function(frame) {
this.$.chartLoader.generateRequest();
}.bind(this),
function(e) {
window.setTimeout(function() {
this.connectWs();
}.bind(this), 2500);
}.bind(this)
);
this.stompClient.subscribe('/topic/chart/' + chart.id,
function(message, headers) {
this.setChartData(chart, JSON.parse(message.body));
}.bind(this), {
"id" : "" + chart.id
}
);
服务器端,我怎样才能在注解的方法当前登录的用户?
@SubscribeMapping("/chart/{id}")
public void subscribeMapping(@DestinationVariable("id") final short id) {
// Here I would need the current user...
messagingTemplate.convertAndSend("/topic/chart/" + id, chartService.getChartData(account, sensor, new Date(), new Date()));
}
我已经试过SecurityContextHolder.getContext().getAuthentication()
,但它返回null
。