Question

我想等到我的过程完成后再返回我的Speechlet响应，否则似乎切断了我的过程，因此没有完成，我实际上认为它可能会冻结该过程，但这不是我的愿望。

我该如何等待？

@Override
public SpeechletResponse onIntent(SpeechletRequestEnvelope<IntentRequest> requestEnvelope) {
    IntentRequest request = requestEnvelope.getRequest();

    Intent intent = request.getIntent();
    String intentName = (intent != null) ? intent.getName() : null;

    if (intentName == null) return null;

    switch (intentName) {
        case IntentTitle.NEW_TICKET:

            switch (request.getDialogState()) {
                case STARTED:
                    return Response.getDialogueResponse(intent, true);
                case IN_PROGRESS:
                    return Response.getDialogueResponse(intent, false);
                case COMPLETED:
                    String numberString = intent.getSlot(SlotTitle.ID).getValue();
                    if (!NumberUtils.isCreatable(numberString)) return Response.ERROR;

                    Member member = Info.GUILD.getMemberById(numberString);
                    User sender = UserDB.getUser(member);

                    CommissionTicket commissionTicket = new CommissionTicket(sender);
                    commissionTicket.create();

                    //wait until processes finish before continuing

                    return Response.NEW_TICKED_CREATED;
            }
    }

    return null;
}

更新：看过CloudWatch日志，好吧，几乎我正在发生的事情正在发生……看看这些日志的时间（我将它们运行了3次不同的时间，因此记录了3次不同的时间，但是除了相同的lambda会话）

public void create() {
    System.out.println("log2");
    GuildController guildController = Info.GUILD.getController();

    guildController.createTextChannel(ticketType.name().toLowerCase() + "-" + creator.getName() + "-" + id.value()).queue(channel -> {
        System.out.println("log3");
                ChannelManager channelManager = channel.getManager();
                GuildManager guildManager = channelManager.getGuild().getManager();
                List<Member> members = guildManager.getGuild().getMembers();
                List<Member> admins = new ArrayList<>();

                for (Member member : members) {
                    for (Role role : member.getRoles()) {
                        if (!role.getName().equalsIgnoreCase(Info.ADMIN_STRING)) continue;
                        admins.add(member);
                    }
                }
        System.out.println("log4");
                for (Member member : members) {
                    if (member.equals(creator.getMember())) continue;
                    channel.createPermissionOverride(member).setDeny(Permission.MESSAGE_READ).queue();
                }
        System.out.println("log5");

                for (Member admin : admins) {
                    if (admin.equals(creator.getMember())) continue;
                    channel.createPermissionOverride(admin).setAllow(Permission.MESSAGE_READ).queue();
                }
        System.out.println("log6");

                BotMessage botMessage = new BotMessage();
                botMessage
                        .setTitle("New Ticket! User: " + creator.getName())
                        .setColour(Color.CYAN)
                        .setDescription("Please enter your full request here! \n" +
                                "Make sure to let us know whether you are looking for a quote/timeframe,\n" +
                                "or have a budget in mind, and we will work around you!\n\n" +
                                "A sales representative will be with you as soon as possible!")
                        .send((TextChannel) channel);
        System.out.println("log7");

                this.textChannel = (TextChannel) channel;

                TicketDB.addTicket(this);
        System.out.println("log8");
            }
    );

日志：

https://gyazo.com/0ad2baa8d1438be8364dd1112159c4f4 https://gyazo.com/e197f33335046afe3c9f8f1ace267d30

更新实现了Future类，可以运行，但仍然有些问题。现在它可以完全创建票证，这很棒，但是，当我再次发送相同的呼叫时，由于某种原因，它会在创建下一个票证之前在同一通道中发送一条消息。

所以，要模拟...

通过AWS Lambda进行函数调用

完全创建票证

再次通过AWS Lambda函数调用

在上一个票务频道中发送一两个消息

完全创建新票证

https://gyazo.com/dc6e4391f4964f41a73f1c3be92190f9

@Override
public SpeechletResponse onIntent(SpeechletRequestEnvelope<IntentRequest> requestEnvelope) {
    IntentRequest request = requestEnvelope.getRequest();

    Intent intent = request.getIntent();
    String intentName = (intent != null) ? intent.getName() : null;

    if (intentName == null) return null;

    switch (intentName) {
        case IntentTitle.NEW_TICKET:

            switch (request.getDialogState()) {
                case STARTED:
                    return Response.getDialogueResponse(intent, true);
                case IN_PROGRESS:
                    return Response.getDialogueResponse(intent, false);
                case COMPLETED:
                    String numberString = intent.getSlot(SlotTitle.ID).getValue();
                    if (!NumberUtils.isCreatable(numberString)) return Response.ERROR;

                    Member member = Info.GUILD.getMemberById(numberString);
                    User sender = UserDB.getUser(member);

                    System.out.println("log1");

                    Future<Ticket> commissionTicket = new CommissionTicket(sender).create();
                    try {
                        commissionTicket.get(10000, TimeUnit.MILLISECONDS);
                    } catch (Exception e) {
                        e.printStackTrace();
                    }

                    //wait until processes finish before continuing

                    return Response.NEW_TICKED_CREATED;
            }
    }

    return null;
}

在票务类中：

public Future<Ticket> create() {
    System.out.println("log2");
    GuildController guildController = Info.GUILD.getController();
    RequestFuture<Channel> channelRequestFuture = guildController.createTextChannel(ticketType.name().toLowerCase() + "-" + creator.getName() + "-" + id.value()).submit();
    try {
        Channel channel = channelRequestFuture.get(10000, TimeUnit.MILLISECONDS);
                    System.out.println("log3");
        ChannelManager channelManager = channel.getManager();
        GuildManager guildManager = channelManager.getGuild().getManager();
        List<Member> members = guildManager.getGuild().getMembers();
        List<Member> admins = new ArrayList<>();

        for (Member member : members) {
            for (Role role : member.getRoles()) {
                if (!role.getName().equalsIgnoreCase(Info.ADMIN_STRING)) continue;
                admins.add(member);
            }
        }
        System.out.println("log4");
        for (Member member : members) {
            if (member.equals(creator.getMember())) continue;
            channel.createPermissionOverride(member).setDeny(Permission.MESSAGE_READ).queue();
        }
        System.out.println("log5");

        for (Member admin : admins) {
            if (admin.equals(creator.getMember())) continue;
            channel.createPermissionOverride(admin).setAllow(Permission.MESSAGE_READ).queue();
        }
        System.out.println("log6");

        BotMessage botMessage = new BotMessage();
        botMessage
                .setTitle("New Ticket! User: " + creator.getName())
                .setColour(Color.CYAN)
                .setDescription("Please enter your full request here! \n" +
                        "Make sure to let us know whether you are looking for a quote/timeframe,\n" +
                        "or have a budget in mind, and we will work around you!\n\n" +
                        "A sales representative will be with you as soon as possible!")
                .send((TextChannel) channel);
        System.out.println("log7");

        this.textChannel = (TextChannel) channel;

        TicketDB.addTicket(this);
        System.out.println("log8");
        Future<Ticket> future = ConcurrentUtils.constantFuture(this);
        return future;
    } catch (Exception e) {
        e.printStackTrace();
    }

    if (!userIsInTicket(creator)) users.add(creator);
    return null;

}

Answer 1

我不确定commissionTicket.create();内部发生了什么（我想这是您的代码，不是某些库的一部分，并且它以某种方式异步运行），但是一种解决方案是让该方法返回一个{ {1}}对象，然后等待它完成。类似于：

Future<>

如何等待进程完成Java AWS Lambda

1 个答案: