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

考虑使用协程

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

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)

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。

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。

我有三个解决方案:

显式地指定Looper: 处理程序(Looper.getMainLooper ()) .postDelayed ({ / /代码 },持续时间) 指定隐式线程本地行为: 处理程序(Looper.myLooper () ! !) .postDelayed ({ / /代码 },持续时间) 使用线程: 线程({ 尝试{ thread . sleep (3000) } catch (e:异常){ 把e } / /代码 }).start ()