Java Spring异步请求处理 @PutMapping(“/guild/{guildId}/audioplayer/queue”) public WebAsyncTask queueAddRoute(@PathVariable String GuidId,@RequestParam String uri)引发UnknownGuildException{ Guild=shardManager.getGuildById(guildId); if(guild==null){ 抛出新的UnknownGuildException(“未找到帮会”+guildId+); } cachedBuilderData=GlobalCacheHolder.getGuildCache().get(guild.getIdLong()); GuildMusicManager musicMan=data.getMusicManager(); //队列(字符串、使用者成功、使用者错误、布尔中断) musicMan.queue(uri,可播放->{ },例外->{ },假); 返回新的WebAsyncTask(); }
在这个控制器方法中,我想向队列添加一些东西,但是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,以便用户获得消息Java Spring异步请求处理 @PutMapping(“/guild/{guildId}/audioplayer/queue”) public WebAsyncTask queueAddRoute(@PathVariable String GuidId,@RequestParam String uri)引发UnknownGuildException{ Guild=shardManager.getGuildById(guildId); if(guild==null){ 抛出新的UnknownGuildException(“未找到帮会”+guildId+); } cachedBuilderData=GlobalCacheHolder.getGuildCache().get(guild.getIdLong()); GuildMusicManager musicMan=data.getMusicManager(); //队列(字符串、使用者成功、使用者错误、布尔中断) musicMan.queue(uri,可播放->{ },例外->{ },假); 返回新的WebAsyncTask(); },java,spring,asynchronous,Java,Spring,Asynchronous,在这个控制器方法中,我想向队列添加一些东西,但是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,以便用户获得消息 @PutMapping("/guild/{guildId}/audioplayer/queue") public WebAsyncTask<String> queueAddRoute(@PathVariable String guildId, @RequestParam String uri) throws
@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<>();
}