What is the equivalent of javascript setTimeout in

问题:

I need to implement a function to run after 60 seconds of clicking a button. Please help, I used the Timer class, but I think that that is not the best way.

回答1:

"I used the Timer class, but I think that that is not the best way."

The other answers assume you are not using Swing for your user interface (button).

If you are using Swing then do not use Thread.sleep() as it will freeze your Swing application.

Instead you should use a javax.swing.Timer.

See the Java tutorial How to Use Swing Timers and Lesson: Concurrency in Swing for more information and examples.

回答2:

Asynchronous implementation with JDK 1.8:

public static void setTimeout(Runnable runnable, int delay){
    new Thread(() -> {
        try {
            Thread.sleep(delay);
            runnable.run();
        }
        catch (Exception e){
            System.err.println(e);
        }
    }).start();
}

To call with lambda expression:

setTimeout(() -> System.out.println("test"), 1000);

Or with method reference:

setTimeout(anInstance::aMethod, 1000);

To deal with the current running thread only use a synchronous version:

public static void setTimeoutSync(Runnable runnable, int delay) {
    try {
        Thread.sleep(delay);
        runnable.run();
    }
    catch (Exception e){
        System.err.println(e);
    }
}

Use this with caution in main thread – it will suspend everything after the call until timeout expires and runnable executes.

回答3:

Use Java 9 CompletableFuture, every simple:

CompletableFuture.delayedExecutor(5, TimeUnit.SECONDS).execute(() -> {
  // Your code here executes after 5 seconds!
});

回答4:

You can simply use Thread.sleep() for this purpose. But if you are working in a multithreaded environment with a user interface, you would want to perform this in the separate thread to avoid the sleep to block the user interface.

try{
    Thread.sleep(60000);
    // Then do something meaningful...
}catch(InterruptedException e){
    e.printStackTrace();
}

回答5:

Do not use Thread.sleep or it will freeze your main thread and not simulate setTimeout from JS. You need to create and start a new background thread to run your code without stoping the execution of the main thread. Like this:

new Thread() {
    @Override
    public void run() {
        try {
            this.sleep(3000);
        } catch (InterruptedException e) {
            e.printStackTrace();
        }


        // your code here

    }
}.start();

回答6:

public ScheduledExecutorService = ses;
ses.scheduleAtFixedRate(new Runnable(){
    run(){
            //running after specified time
}
}, 60, TimeUnit.SECONDS);

its run after 60 seconds from scheduleAtFixedRate https://docs.oracle.com/javase/7/docs/api/java/util/concurrent/ScheduledExecutorService.html

回答7:

You should use Thread.sleep() method.

try {

    Thread.sleep(60000);
    callTheFunctionYouWantTo();
} catch(InterruptedException ex) {

}

This will wait for 60,000 milliseconds(60 seconds) and then execute the next statements in your code.

回答8:

There is setTimeout() method in underscore-java library.

Code example:

import com.github.underscore.U;
import com.github.underscore.Function;

public class Main {

    public static void main(String[] args) {
        final Integer[] counter = new Integer[] {0};
        Function<Void> incr = new Function<Void>() { public Void apply() {
            counter[0]++; return null; } };
        U.setTimeout(incr, 100);
    }
}

The function will be started in 100ms with a new thread.

回答9:

Using the java.util.Timer:

new Timer().schedule(new TimerTask() {
    @Override
    public void run() {
        // here goes your code to delay
    }
}, 300L); // 300 is the delay in millis

Here you can find some info and examples.