如何修复此代码中的弃用警告?或者,还有其他的选择吗?

Handler().postDelayed({
    context?.let {
        //code
    }
}, 3000)

当前回答

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。

其他回答

使用Executor而不是handler获取更多信息。 使用ScheduledExecutorService实现post delay:

ScheduledExecutorService worker = Executors.newSingleThreadScheduledExecutor();
Runnable runnable = () -> {
    public void run() {
        // Do something
    }
};
worker.schedule(runnable, 2000, TimeUnit.MILLISECONDS);

被弃用的函数是Handler的构造函数。请改用Handler(loop . mylooper ()) .postDelayed(runnable, delay)

使用生命周期范围会更容易。内部活动或片段。

 lifecycleScope.launch {
     delay(2000)
     // Do your stuff
 }

或者使用处理器

        Handler(Looper.myLooper()!!)
import android.os.Looper
import android.os.Handler

inline fun delay(delay: Long, crossinline completion: () -> Unit) {
    Handler(Looper.getMainLooper()).postDelayed({
        completion()
    }, delay)
}

例子:

delay(1000) {
    view.refreshButton.visibility = View.GONE
}

考虑使用协程

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