Spring异步请求处理

Question

    @PutMapping("/guild/{guildId}/audioplayer/queue")
    public WebAsyncTask<String> queueAddRoute(@PathVariable String guildId, @RequestParam String uri) throws UnknownGuildException {
        Guild guild = shardManager.getGuildById(guildId);
        
        if(guild == null) {
            throw new UnknownGuildException("Guild " + guildId + " was not found");
        }
        
        CachedGuildData data = GlobalCacheHolder.getGuildCache().get(guild.getIdLong());
        GuildMusicManager musicMan = data.getMusicManager();
        
        //musicMan.queue(String, Consumer<Playable> success, Consumer<Exception> error, boolean interrupt)
        musicMan.queue(uri, playable -> {
            
        }, exception -> {
            
        }, false);
        
        return new WebAsyncTask<>();
    }

在此Controller方法中，我想向队列中添加一些内容，但是该队列返回两个使用者，一个用于成功，一个用于异常，我如何返回异常或成功使用者，以便用户得到一条消息