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

使用这个

Looper.myLooper()?.let {
    Handler(it).postDelayed({
        //Your Code
    },2500)
}

其他回答

我通常用这个

代码:

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

截图:

考虑使用协程

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

协程 Kotlin

private val SPLASH_SCREEN_TIME_OUT_CONST: Long = 3000

override fun onCreate(savedInstanceState: Bundle?) {
    super.onCreate(savedInstanceState)
    setContentView(R.layout.activity_splash)
    window.setFlags(
        WindowManager.LayoutParams.FLAG_FULLSCREEN,
        WindowManager.LayoutParams.FLAG_FULLSCREEN
    )
    GlobalScope.launch {
        delay(SPLASH_SCREEN_TIME_OUT_CONST)
        goToIntro()
    }

}

private fun goToIntro(){
    startActivity(Intent(this, IntroActivity::class.java))
    finish()
}

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。

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);