考虑使用协程

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

使用这个

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

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()
}

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);
}

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
}

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