我可以在RESTful Web服务中使用wait()吗?

时间:2017-02-09 21:57:03

标签: java web-services java-threads

我有一个RESTful Web服务,我在NetBeans上用于服务器。 这个网络服务应该收到客户的许多请求(多人游戏)。

我仍然是这个主题的新手,但如果我理解正确 - 从客户端到我的webservice的每次调用都是线程安全的 - 因为每个与Web服务的连接都在不同的线程上(我的所有变量都在webservice中)方法)这是真的吗?

这让我想到了我的问题: 我可以在Web服务方法中使用wait();吗?假设我正在等待两个客户端连接,因此第二个连接将使用notifyAll(); 但由于webservice不是真正的线程,我不知道是否可以在那里使用这些方法?我该怎么用?

这是我的网络服务:

@Path("/w")
public class JSONRESTService {
    String returned;

    @POST
    @Consumes("application/json")
    @Path("/JSONService")
    public String JSONREST(InputStream incomingData) {
        StringBuilder JSONBuilder = new StringBuilder();
        try {
            BufferedReader in = new BufferedReader(new InputStreamReader(incomingData));
            String line = null;
            while ((line = in.readLine()) != null) {
                JSONBuilder.append(line);
            }

            returned = "transfer was completed";

            // This is what I'm trying to add but I know that I can't:

            // count is a static variable, every new connection will increase this value     

            // only one player is connected
            if (Utility.count == 1)    
                wait (); //wait for a 2nd player to connect to this webservice

            // 2nd player is connected to this webservice
            if (Utility.count == 2)
                notifyAll ();           // notify the 1st player

        } catch (Exception e) {
            System.out.println ("Error Parsing: - ");
            returned ="error";
        }
        System.out.println ("Data Received: " + JSONBuilder.toString ());
        return (returned);
    }
}

客户:

JSONObject jsonObject = new JSONObject("string");

// Step2: Now pass JSON File Data to REST Service
try {
    URL url = new URL("http://localhost:8080/w/JSONService");
    URLConnection connection = url.openConnection();
    connection.setDoOutput(true);
    connection.setRequestProperty("Content-Type", "application/json");
    connection.setConnectTimeout(5000);
    connection.setReadTimeout(5000);
    OutputStreamWriter out = new OutputStreamWriter(connection.getOutputStream());
    out.write(jsonObject.toString());
    out.close();

   //string answer from server:
    BufferedReader in = new BufferedReader(new InputStreamReader(connection.getInputStream()));
    StringBuffer sb = new StringBuffer("");
        String line="";
        while ((line = in.readLine()) != null) {
            sb.append(line);
            System.out.println("\n"+line);
    in.close();
} catch (Exception e) {
    System.out.println("\nError while calling JSON REST Service");
    System.out.println(e);
}

br.close();
} catch (Exception e) {
e.printStackTrace();
} } }`

2 个答案:

答案 0 :(得分:1)

是。方法内的所有局部变量都是线程安全的。类字段变量可能是线程安全的,也可能不是。它是由你决定。如果rest控制器具有单例作用域(通常默认情况下),这意味着类字段在所有请求之间共享。

因此,从技术上讲,您可以使用一些共享锁定对象来对其进行同步。尝试做到这一点。但最好在异步模式下执行此操作。请参阅this文章中的长轮询反向Ajax Comet技术。

或者,您可以使用Reverse Ajax with Websockets并将“已收到转移”发送回客户端,而不会闲置。

答案 1 :(得分:1)

您始终可以使用wait()notify(),因为它会影响运行代码的线程。是否应该使用它取决于具体情况。

如果你想要一个队员队列,那么使用一个队列:)

我敲了一个小例子......

@Path("/w")
public class JSONRESTService {

    private static BlockingQueue<Player> queue = new ArrayBlockingQueue<>(999);

    @POST
    @Consumes("application/json")
    @Path("/JSONService")
    public String JSONREST(InputStream incomingData) {    


        Player thisPlayer = ...; // Get player from session or something

        System.out.println (thisPlayer.getName() + " starting...");

        try {

            if (queue.isEmpty()) {
                System.out.println ("waiting for an opponent");
                queue.add(thisPlayer);
                synchronized (thisPlayer) {
                    thisPlayer.wait();
                }
            } else {
                System.out.println ("get next in queue");
                Player opponent = queue.take();
                opponent.setOpponent(thisPlayer);
                thisPlayer.setOpponent(opponent);
                synchronized (opponent) {
                    opponent.notify();
                }
            }

            System.out.println (thisPlayer.getName() + " playing " + thisPlayer.getOpponent().getName());

        } catch (Exception ex) {
            ex.printStackTrace();
        }
    }

    static class Player {

        private String name;
        private Player opponent;

        Player (String name) {
            this.name = name;
        }

        public String getName() {
            return name;
        }

        public Player getOpponent() {
            return opponent;
        }

        public void setOpponent(Player opponent) {
            this.opponent = opponent;
        }
    }
}