I have a simple RabbitMQ test program randomly enqueuing messages, and another reading them, all using Spring-AMQP. If the consumer dies (for example killing a process without having a chance to close its connection or channel), any messages that it had not acknowledged appear to remain unacknowledged forever. I have seen a number of references (for example this question) that say that the cha
我有一个简单的RabbitMQ测试程序随机排队消息，另一个阅读它们，全部使用Spring-AMQP。 如果消费者死亡（例如，在没有机会关闭其连接或通道的情况下杀死一个进程），则任何未被确认的消息似乎永远都不会被确认。 我见过很多参考资料（例如这个问题），说道这个频道在没有连接的情况下就会死亡，剩下的未分类的消息将被重新发送。 这不是我看到的行为 - 相反，我获得了越来越多的标记为IDLE的频道列表，以及越来越多的标记为
I've googled around everywhere, but cannot find much for rendering strings to textures and then displaying that texture on a quad on the screen. Can someone provide a run-down on the process or provide good resources that describe how? Is rendering strings to textures even the best method for displaying text in an Android OpenGL ES app? EDIT: Okay, so LabelMaker interferes with alpha blen
我到处都是搜索引擎，但找不到很多东西来渲染字符串到纹理，然后在屏幕上的四边形上显示纹理。 有人可以在流程中提供破解或提供描述如何的好资源吗？ 即使是在Android OpenGL ES应用程序中显示文本的最佳方法，是否将字符串呈现为纹理？ 编辑：好吧，所以LabelMaker干涉alpha混合，纹理（从具有透明背景的PNG创建）现在有一个纯黑色的背景，而不是一个透明的背景。 如果我注释掉所有与LabelMaker相关的代码，它就可以正常工
I have a base class that captures some functionality common to two classes. In other words, I can create one base class and make these two classes subclasses of that base class. However, for each of these sub classes, the number of instances can be created is 1 (ie each sub class has to be a singleton). I googled and found that there's a reasonable debate going on over this. Although ther
我有一个基类捕捉两个类通用的一些功能。 换句话说，我可以创建一个基类，并且创建这个基类的这两个类的子类。 但是，对于这些子类中的每一个，可以创建的实例数为1（即每个子类都必须是单例）。 我搜索了一下，发现这方面正在进行一场合理的辩论。 虽然有几种解决方案可用，但我不确定它们是否适合我的情况。 任何人都可以告诉我我应该如何设计这个？ 您可以将每个类分别制作为单例，并使基类变为抽象。 不知道什么是
My Mac currently is running Snow Leopard (10.6.8), and I would like to develop with Java 7. I downloaded the Java 7 OS X installer from Oracle's website, started to run it. Unfortunately, I was immediately greeted with a message that said the installer is supported only on OS X Lion (10.7.3). Is there any way I can get Java 7 on my machine with Snow Leopard? I am not buying a new OS just t
我的Mac目前运行Snow Leopard（10.6.8），我想用Java 7进行开发。我从Oracle网站下载了Java 7 OS X安装程序，开始运行它。 不幸的是，我立即迎来了一条消息，称只有OS X Lion（10.7.3）支持安装程序。 有什么方法可以在我的机器上使用Snow Leopard获得Java 7？ 我不是为了升级我的Java而购买新的操作系统。 我搜索了一些，但我没有找到太多。 我希望我能在这里找到一些帮助。 谢谢。 使用http://www.charlessoft.com/中的P
I was trying to send sms using java and for that I first tried with thilio.But I got error which is mentioned at the bottom.Then i downloaded,created an account in nexmo but again i get the same error.Please anybody those who had already worked on the above two can tell me why this issue come. import com.nexmo.messaging.sdk.NexmoSmsClient; import com.nexmo.messaging.sdk.SmsSubmissionResult; impo
我正在尝试使用java发送短信，为此我首先尝试了thilio.But然后我得到了底部提到的错误。然后我下载，在nexmo中创建了一个帐户，但是我又得到了同样的错误。请任何人已经在上面的两个工作可以告诉我为什么会出现这个问题。 import com.nexmo.messaging.sdk.NexmoSmsClient; import com.nexmo.messaging.sdk.SmsSubmissionResult; import com.nexmo.messaging.sdk.messages.TextMessage; public class SendTextMessage { pub
I have an email system in my website and as of now there is no way to show notification to user in my website. Until unless user login to email client they will not come to know that they have received a message. Can i accomplish this with the help of AWS SQS and SNS? As this a website, there are many registered user. They will be sending message to any user and say when the recipient of the
我的网站上有一个电子邮件系统，截至目前无法在我的网站上向用户显示通知。 除非用户登录到电子邮件客户端，否则他们不会知道他们已收到消息。 我可以在AWS SQS和SNS的帮助下完成此操作吗？ 作为这个网站，有很多注册用户。 他们会发送消息给任何用户，并说当电子邮件的收件人是活动并登录到网站，我需要显示通知。 如果用户离线，我只需要在登录页面的页眉上显示这些消息。 我怎样才能做到这一点？ 我尝试在SQS管理控
In my game there are several class I have written including room, lamp, chest, java, player, key and map. These have all been tested and are correct so now I am writing my adventure class which is the driver for the program. I need to set the players room location to [0][0] and i can't figure out how. here is what i have so far in my room and adventure class. public class Adventure {
在我的游戏中，我写了几个课程，包括房间，灯，胸部，Java，播放器，键和地图。 这些都经过了测试，是正确的，所以现在我正在写我的冒险课，这是该计划的驱动程序。 我需要将玩家房间的位置设置为[0] [0]，我无法弄清楚如何。 这是我到目前为止在我的房间和冒险课上所拥有的。 public class Adventure { Scanner in = new Scanner(System.in); private Room room; public Adventure() { Player player
I need to read the heart rate of the user at that instant. I tried using this code located in my main activity in the Wear project. public class MainActivity extends WearableActivity implements SensorEventListener { private static final String TAG = "MainActivity"; private TextView mTextViewHeart; SensorManager mSensorManager; Sensor mHeartRateSensor; SensorEventListener s
我需要在当时读取用户的心率。 我尝试使用位于Wear项目主要活动中的代码。 public class MainActivity extends WearableActivity implements SensorEventListener { private static final String TAG = "MainActivity"; private TextView mTextViewHeart; SensorManager mSensorManager; Sensor mHeartRateSensor; SensorEventListener sensorEventListener; @Override protected void onCreate
I have a problem, I want to click on the list, calling a new activity and rename the button to another name. I tried several things, nothing worked, can someone please help me? My class EditarTimes private AdapterView.OnItemClickListener selecionarTime = new AdapterView.OnItemClickListener() { public void onItemClick(AdapterView arg0, View arg1, int pos, long id) { t = t
我遇到了问题，我想单击列表，调用一个新的活动并将该按钮重命名为另一个名称。 我尝试了几件事，没有任何工作，有人可以帮我吗？ 我的课EditarTimes private AdapterView.OnItemClickListener selecionarTime = new AdapterView.OnItemClickListener() { public void onItemClick(AdapterView arg0, View arg1, int pos, long id) { t = times.get(pos); CadastroTimes cad = new Cadast
public abstract class Beverage { protected String Description; public String getDescription(){ return Description; } public abstract int cost(); } public class Espresso extends Beverage{ public int cost(){ return 2; } public Espresso(){ Description = "Espresso"; } } abstract class CondimentDecorator extends Beverage{ public
public abstract class Beverage { protected String Description; public String getDescription(){ return Description; } public abstract int cost(); } public class Espresso extends Beverage{ public int cost(){ return 2; } public Espresso(){ Description = "Espresso"; } } abstract class CondimentDecorator extends Beverage{ public