我已经阅读了关于Servlet 线程处理的精彩文章。
我有以下问题:我想创建一个简单的 Servlet,它启动一个在第一个版本中生成随机消息的新线程,或者 - 根据参数 - 发送一个包含自上次请求以来产生的所有消息的响应。
我在浏览器站点上使用 JQuery AJAX 调用处理超时请求。
当我运行接收器调用时,我只收到自线程同时崩溃以来产生的第一条消息。看起来这将是一个线程安全问题,如上述文章中所述,但我可以准确地弄清楚。日志给了我以下信息:
SEVERE: Exception in thread "Thread-75"
SEVERE: java.lang.IllegalMonitorStateException
at java.lang.Object.wait(Native Method)
at com.lancom.lsr.util.RandomMessageProducer.run(RandomMessageProducer.java:35)
at java.lang.Thread.run(Thread.java:722)
SEVERE: at java.lang.Object.wait(Native Method)
SEVERE: at com.lancom.lsr.util.RandomMessageProducer.run(RandomMessageProducer.java:35)
SEVERE: at java.lang.Thread.run(Thread.java:722)
这是我当前的 servlet 代码:
public class MyServlet extends HttpServlet {
...
private RandomMessageProducer rmProducer;
private Thread rmpThread;
...
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
Map<Integer, String> msg = new HashMap<Integer, String>();
Gson gson = new Gson();
String sDevs = request.getParameter("devices"); // Option 1
String sGetMsg = request.getParameter("rec_msg"); // Option 2
PrintWriter pw = response.getWriter();
// Request: Send information and run thread
if (sDevs != null && STATUS==0) {
/* Start a dummy producer */
rmProducer = new RandomMessageProducer();
rmpThread = new Thread( rmProducer )
rmpThread.start();
pw.print("{\"1\": \"Action started!\"}");
STATUS=1;
}
// Request: Receive messages
else if (sGetMsg != null) {
List<String> logs = rmProducer.getLogMsg();
for (String lmsg : logs) {
// Check if we can update the current Status
if (msg.equals("<<<FIN_SUCC>>>") || msg.equals("<<<FIN_ERR>>>")) {
STATUS=0;
}
msg.put(0, lmsg);
}
String json = gson.toJson(msg);
pw.print(json);
}
pw.close();
}
}
这是我简单的消息生产者线程:
public class RandomMessageProducer implements Runnable {
private Queue<String> msgQueue = new LinkedList<String>();
@Override
public void run() {
Random randomGenerator = new Random();
for (int idx = 1; idx <= 100; ++idx){
int randomInt = randomGenerator.nextInt(100);
msgQueue.add("Generated : " + randomInt);
try {
wait(500);
} catch (InterruptedException e) {
msgQueue.add("<<<FIN_ERR>>>");
e.printStackTrace();
}
}
msgQueue.add("<<<FIN_SUCC>>>");
}
public List<String> getLogMsg() {
List<String> res = new ArrayList<String>();
while (!msgQueue.isEmpty()) {
res.add( msgQueue.poll() );
}
return res;
}
}
请求全部执行 1000 毫秒
您可能看到我的推理错误吗?
非常感谢!