考虑使用协程

scope.launch {
    delay(3000L)
    // do stuff
}

如果你想避免在Kotlin (?或者!!)你可以使用loop . getmainlooper()如果你的Handler正在处理一些UI相关的事情，像这样:

Handler(Looper.getMainLooper()).postDelayed({
   Toast.makeText(this@MainActivity, "LOOPER", Toast.LENGTH_SHORT).show()
}, 3000)

注意:如果使用fragment，请使用requireContext()而不是this@MainActivity。

我通常用这个

代码:

Handler(Looper.myLooper() ?: return).postDelayed({
           // Code what do you want
        }, 3000)

截图:

根据文档(https://developer.android.com/reference/android/os/Handler#Handler()):

Implicitly choosing a Looper during Handler construction can lead to bugs where operations are silently lost (if the Handler is not expecting new tasks and quits), crashes (if a handler is sometimes created on a thread without a Looper active), or race conditions, where the thread a handler is associated with is not what the author anticipated. Instead, use an Executor or specify the Looper explicitly, using Looper#getMainLooper, {link android.view.View#getHandler}, or similar. If the implicit thread local behavior is required for compatibility, use new Handler(Looper.myLooper()) to make it clear to readers.

我们应该停止使用没有Looper的构造函数，而是指定一个Looper。

Java的答案

我写了一个容易使用的方法。您可以在项目中直接使用此方法。delayTimeMillis可以是2000，这意味着该代码将在2秒后运行。

private void runJobWithDelay(int delayTimeMillis){
    new Handler(Looper.getMainLooper()).postDelayed(new Runnable() {
        @Override
        public void run() {
            //todo: you can call your method what you want.
        }
    }, delayTimeMillis);
}

如果你使用变量作为处理程序和可运行的，然后像这样使用它。

private Handler handler;
private Runnable runnable;

handler = new Handler(Looper.getMainLooper());
    handler.postDelayed(runnable = () -> {
        // Do delayed stuff here
         handler.postDelayed(runnable, 1000);
    }, delay);

你还需要删除onDestroy()中的回调

@Override
public void onDestroy() {
    super.onDestroy();
    if (handler != null) {
        handler.removeCallbacks(runnable);
    }
}