如何等待多个服务完成？

Question

我正在运行这样的多个服务：（例如，在多个线程中读取文件）

for (int i = 0; i < 3; i++) {
        ReadService readService = new ReadService();
        readService.start();
}

//wait until all services have been completed
System.out.println("All services done!");

ReadService是一个扩展Service类的类，它正在做一些事情，比如读取文件。它是从另一个线程调用的，该线程不是JavaFX应用程序线程。

我如何等到所有这些服务完成后再调用系统。出来println？

可复制的例子：

import javafx.application.Application;
import javafx.concurrent.Service;
import javafx.concurrent.Task;
import javafx.stage.Stage;

public class HelloApplication extends Application {

    @Override
    public void start(Stage stage) {
        for (int i = 0; i < 3; i++) {
            ReadService readService = new ReadService();
            readService.start();
        }
        // wait until all services have been completed
        System.out.println("Services done");
        // continue with some other code, for example check the read file on anything
    }

    public static void main(String[] args) {
        launch();
    }
}

class ReadService extends Service<Boolean> {

    @Override
    protected Task<Boolean> createTask() {
        return new Task<>() {
            @Override
            protected synchronized Boolean call() throws Exception {
                // do something, for example read a file
                System.out.println("wait...");
                wait(5000);
                System.out.println("waiting done");
                return null;
            }
        };
    }
}

Answer 1

您正在从FX Application Thread（这是执行start（）方法的地方）启动服务，并且您不能阻止该线程。所以一种方法是为完成的服务数量创建一个计数器，并在它达到零时做出响应。

请注意，这里的所有新内容（创建和更新servicesPending属性，以及在服务完成时执行的代码）都在FX Application Thread上执行，所以如果您在服务完成时更新UI，这种方法是合适的...

@Override
public void start(Stage stage) {

    int numServices = 3 ;

    IntegerProperty servicesPending = new SimpleIntegerProperty(numServices);

    servicesPending.addListener((obs, oldValue, newValue) -> {
         if (newValue == 0) {
             // code to execute when all services are complete
             System.out.println("Services done");
         }
    });

    for (int i = 0; i < numServices; i++) {
        ReadService readService = new ReadService();
        readService.setOnSucceeded(e -> servicesPending.set(servicesPending.get() - 1));
        readService.start();
    }
    
}

另一方面，如果您在服务完成时所做的工作与UI无关，那么您可以创建一个新的线程，该线程将阻止服务完成，然后在该后台线程上执行该工作。实现这一点的一种方法是使用CountDownLatch：

@Override
public void start(Stage stage) {

    int numServices = 3 ;

    CountDownLatch latch = new CountDownLatch(numServices);

    for (int i = 0; i < numServices; i++) {
        ReadService readService = new ReadService();
        readService.setOnSucceeded(e -> latch.countDown());
        readService.start();
    }

    Thread onServicesCompleted = new Thread(() -> {
        try {
            latch.await();
        } catch(InterruptedException exc) {
            Thread.currentThread().interrupt();
        }
        System.out.println("Services done");
        // other work to do when services are complete...
    });
    onServicesCompleted.start();
    
}

@jewelsea在评论中提出了类似的解决方案。如果使用Tasks而不是Services，则可以调用get（），该函数将一直阻塞，直到任务完成：

@Override
public void start(Stage stage) {

    int numServices = 3 ;

    List<Task<Boolean>> tasks = new ArrayList<>();
    ExecutorService exec = Executors.newCachedThreadPool();


    for (int i = 0; i < numServices; i++) {
        ReadService readService = new ReadService();
        tasks.add(readService);
    }

    exec.invokeAll(tasks);

    Task<Void> onServicesCompleted = new Task<>() {
        @Override
        protected Void call() throws Exception {
            for (Task<Boolean> task : tasks) {
                task.get();
            }
            System.out.println("Services Done");
            // other work to be done...
        }
    };
    exec.execute(onServicesCompleted);

}

和

class ReadService extends Task<Boolean> {
    @Override
    protected synchronized Boolean call() throws Exception {
        // do something, for example read a file
        System.out.println("wait...");
        wait(5000);
        System.out.println("waiting done");
        return null;
    }
}

如果您想在单个ReadService都完成时执行更多后台工作，然后希望在此之后执行UI工作，则此解决方案非常好（只需使用onServicesCompleted.setOn成功（...））。